Download 1 On Geometric Proofs: Base angles of an isosceles trapezoid are

On Geometric Proofs:
Base angles of an isosceles trapezoid are equal
Perpendicular bisectors of a triangle meet at a common point
Before demonstrating the above proofs, we should review what sort of geometric tools are
available to us. Not all of what is presented below will be used but it’s useful to know what’s in
the arsenal.
1.
Triangle Congruence Relations: There are four: side-angle-side or SAS (Prop I.4) sideside-side or SSS (Prop I.8), angle-side-angle or ASA and angle-angle-side or AAS (Prop 1.26).
Corresponding sides and angles of congruent triangles are equal.
2.
Constructions I: Using straight edge and compass it is possible to
Bisect an angle (Prop, I.9)
Bisect a finite line segment (Prop I.10)
Construct a perpendicular line from any point on a given line (Prop I.11) or any point not
on a given line (Prop I.12)
3.
Vertical Angles (Prop I.15). If two lines cross the opposite (vertical) angles are equal.
That is 1 = 2
2
1
4.
Parallel Lines: If two parallel lines are crossed by a third, the alternate angles are equal,
1 = 2 , the exterior angle equal to the interior and opposite angle on the same side, 1 = 3 ,
and the two interior angles on the same side equal two right angles, 2 = 4 (Prop I.29).
1
4
2
3
In addition if two lines are crossed by a third such that the alternate angles are equal 1 = 2 or
the exterior angle is equal to the interior and opposite angle on the same side 1 = 3 or the two
interior angles on the same side equal two right angles 2 = 4 , then the lines are parallel (Prop
I.27 and I.28)
5.
Parallel Line Construction: Given a line and point not on the line it is possible to
construct a second line through the point that is parallel to the first line. (Prop I.31)
6.
Isosceles triangle result: The base angles of an isosceles triangle are equal (Prop I.5). If
the base angles of a triangle are equal then the triangle is isosceles (Prop I.6)
1
Theorem: The base angles of an isosceles trapezoid are equal
Introduction: A famous theorem of geometry proves the result
that the base angles of an isosceles triangle are equal. A similar
result holds for an isosceles trapezoid; that is in a trapezoid where
the lengths of the two non-parallel sides are equal (and the
parallel sides are of different length) the base angles are equal
Given an isosceles trapezoid ABCD where AD and BC are
parallel and AB equals CD, the base angles BAD and CDA are
equal. Assume BC is the shorter of the two parallel lines.
C
B
1
5
A
6 4
State what is being
proved
Background/
Introduction
Proposition (to be
proved)
3
2
7
D
E
Proof: Begin by constructing a line BE parallel to CD. Construct
the line CE.
Because diagonal CE intersects parallel lines BC and AD angle
BCE (angle 1) equals angle CED (angle 2).
Proof/Argument
Likewise because CE intersects parallel lines BE and CD angle
ECD (angle 3) equals angle CEB (angle 4).
Therefore since CE is common to both triangles ∆BEC and
∆DCE , triangles ∆BEC and ∆DCE are congruent by ASA.
Therefore BE equals CD and since CD equals AB, ∆ABE is
isosceles so angle BAE (angle 5) equals angle BEA (angle 6)
Since AD intersects parallel lines BE and CD angle BEA (angle 6)
equals angle CDA (angle 7).
Since angle BAD (angle 5) also equals angle BEA (angle 6),
angle BAD (angle 5) equals angle CDA (angle 7) and the result
follows.
Conclusion: Therefore the base angles of an isosceles trapezoid
are equal.
Conclusion/Summary
QED.
Corollary: Angle ABC equals angle DCB.
2
The purpose in writing “good mathematics” is to clearly and effectively communicate a
mathematical argument. I generally look for 4 things: clarity, correctness, conciseness,
completeness.
Clarity – This is the “biggie”: obviously any argument presented should be clear and easy to
understand. This is not always easy to do since mathematics can be difficult. Making a difficult
argument clear often presents a challenge; imagination is required. Presenting a good example or
examining a simple case often helps.
Correctness – obviously the logic of the argument can have no holes or flaws; this is the dealbreaker!
Conciseness – Wordiness of argument dulls the reader. An argument presented in 50 words is
easier to understand than an argument in 500 words
Completeness – Did you leave anything out? Did you tie off all the loose ends? Did you forget
something? Leave out an important statement necessary to make the whole argument work?
Completeness is the counter-balance to conciseness and effective writing has the right mix of
both.
A good argument should flow naturally – inexorably – to its conclusion.
The details of how you accomplish the above vary according to the assignment; there is no
universal template when presenting a mathematical argument (or proof). However I have found
that a good mathematical argument has four parts:
a statement (proposition/theorem) of what is to be proved or shown or argued,
an introduction/background which introduces the reader to what is to come,
the “proof” itself (the longest part of the write up) where all the work is done and finally
a short summary/conclusion which brings the whole write up to closure.
As an example the above is a (rather over done but effective) proof that the base angles of an
isosceles triangle are equal. Not all proofs can or need to be done this way but it illustrates one
way to structure a mathematical proof that is clear, correct, concise and complete.
We continue with a second proof that the perpendicular bisectors of a triangle meet at a common
point.
3
Theorem: The perpendicular bisectors of a triangle meet at a
common point.
Introduction: It’s not difficult to see that any two
perpendicular bisectors meet at a common point. We’ll show
that a line from the 3rd side to that common point is also a
perpendicular bisector
State what is being proved
A good introduction sets
up the reader for what is to
follow
Proof: Given triangle ABD
B
D
E
O
A
C
F
Let points D, E, and F be the bisectors of AB, BC, and AC
respectively.
Note the step by step
structure of the proof –
each statement naturally
following from the
previous. A good proof
should naturally flow
Construct the perpendicular bisectors DG and EH that meet at
point O.
By SAS triangle BDO is congruent to triangle ADO so AO
equals BO.
Again by SAS triangle BEO is congruent to triangle CEO so
BO equals CO.
Therefore AO equals CO
Therefore by SSS triangle AFO is congruent to triangle CFO
and it follows that angles AFO and CFO are congruent.
Since these two angles AFO and CFO form a straight line,
they are both right.
Therefore FO is a perpendicular bisector of AC
Conclusion: The perpendicular bisectors of a triangle meet at
a common point - QED
4
The Conclusion brings
closure.
On Geometric Proofs:
Base angles of an isosceles trapezoid are equal
Perpendicular bisectors of a triangle meet at a common point
Before demonstrating the above proofs, we should review what sort of geometric tools are
available to us. Not all of what is presented below will be used but it’s useful to know what’s in
the arsenal.
1.
Triangle Congruence Relations: There are four: side-angle-side or SAS (Prop I.4) sideside-side or SSS (Prop I.8), angle-side-angle or ASA and angle-angle-side or AAS (Prop 1.26).
Corresponding sides and angles of congruent triangles are equal.
2.
Constructions I: Using straight edge and compass it is possible to
Bisect an angle (Prop, I.9)
Bisect a finite line segment (Prop I.10)
Construct a perpendicular line from any point on a given line (Prop I.11) or any point not
on a given line (Prop I.12)
3.
Vertical Angles (Prop I.15). If two lines cross the opposite (vertical) angles are equal.
That is 1 = 2
2
1
4.
Parallel Lines: If two parallel lines are crossed by a third, the alternate angles are equal,
1 = 2 , the exterior angle equal to the interior and opposite angle on the same side, 1 = 3 ,
and the two interior angles on the same side equal two right angles, 2 = 4 (Prop I.29).
1
4
2
3
In addition if two lines are crossed by a third such that the alternate angles are equal 1 = 2 or
the exterior angle is equal to the interior and opposite angle on the same side 1 = 3 or the two
interior angles on the same side equal two right angles 2 = 4 , then the lines are parallel (Prop
I.27 and I.28)
5.
Parallel Line Construction: Given a line and point not on the line it is possible to
construct a second line through the point that is parallel to the first line. (Prop I.31)
6.
Isosceles triangle result: The base angles of an isosceles triangle are equal (Prop I.5). If
the base angles of a triangle are equal then the triangle is isosceles (Prop I.6)
1
Theorem: The base angles of an isosceles trapezoid are equal
Introduction: A famous theorem of geometry proves the result
that the base angles of an isosceles triangle are equal. A similar
result holds for an isosceles trapezoid; that is in a trapezoid where
the lengths of the two non-parallel sides are equal (and the
parallel sides are of different length) the base angles are equal
Given an isosceles trapezoid ABCD where AD and BC are
parallel and AB equals CD, the base angles BAD and CDA are
equal. Assume BC is the shorter of the two parallel lines.
C
B
1
5
A
6 4
State what is being
proved
Background/
Introduction
Proposition (to be
proved)
3
2
7
D
E
Proof: Begin by constructing a line BE parallel to CD. Construct
the line CE.
Because diagonal CE intersects parallel lines BC and AD angle
BCE (angle 1) equals angle CED (angle 2).
Proof/Argument
Likewise because CE intersects parallel lines BE and CD angle
ECD (angle 3) equals angle CEB (angle 4).
Therefore since CE is common to both triangles ∆BEC and
∆DCE , triangles ∆BEC and ∆DCE are congruent by ASA.
Therefore BE equals CD and since CD equals AB, ∆ABE is
isosceles so angle BAE (angle 5) equals angle BEA (angle 6)
Since AD intersects parallel lines BE and CD angle BEA (angle 6)
equals angle CDA (angle 7).
Since angle BAD (angle 5) also equals angle BEA (angle 6),
angle BAD (angle 5) equals angle CDA (angle 7) and the result
follows.
Conclusion: Therefore the base angles of an isosceles trapezoid
are equal.
Conclusion/Summary
QED.
Corollary: Angle ABC equals angle DCB.
2
The purpose in writing “good mathematics” is to clearly and effectively communicate a
mathematical argument. I generally look for 4 things: clarity, correctness, conciseness,
completeness.
Clarity – This is the “biggie”: obviously any argument presented should be clear and easy to
understand. This is not always easy to do since mathematics can be difficult. Making a difficult
argument clear often presents a challenge; imagination is required. Presenting a good example or
examining a simple case often helps.
Correctness – obviously the logic of the argument can have no holes or flaws; this is the dealbreaker!
Conciseness – Wordiness of argument dulls the reader. An argument presented in 50 words is
easier to understand than an argument in 500 words
Completeness – Did you leave anything out? Did you tie off all the loose ends? Did you forget
something? Leave out an important statement necessary to make the whole argument work?
Completeness is the counter-balance to conciseness and effective writing has the right mix of
both.
A good argument should flow naturally – inexorably – to its conclusion.
The details of how you accomplish the above vary according to the assignment; there is no
universal template when presenting a mathematical argument (or proof). However I have found
that a good mathematical argument has four parts:
a statement (proposition/theorem) of what is to be proved or shown or argued,
an introduction/background which introduces the reader to what is to come,
the “proof” itself (the longest part of the write up) where all the work is done and finally
a short summary/conclusion which brings the whole write up to closure.
As an example the above is a (rather over done but effective) proof that the base angles of an
isosceles triangle are equal. Not all proofs can or need to be done this way but it illustrates one
way to structure a mathematical proof that is clear, correct, concise and complete.
We continue with a second proof that the perpendicular bisectors of a triangle meet at a common
point.
3
Theorem: The perpendicular bisectors of a triangle meet at a
common point.
Introduction: It’s not difficult to see that any two
perpendicular bisectors meet at a common point. We’ll show
that a line from the 3rd side to that common point is also a
perpendicular bisector
State what is being proved
A good introduction sets
up the reader for what is to
follow
Proof: Given triangle ABD
B
D
E
O
A
C
F
Let points D, E, and F be the bisectors of AB, BC, and AC
respectively.
Note the step by step
structure of the proof –
each statement naturally
following from the
previous. A good proof
should naturally flow
Construct the perpendicular bisectors DG and EH that meet at
point O.
By SAS triangle BDO is congruent to triangle ADO so AO
equals BO.
Again by SAS triangle BEO is congruent to triangle CEO so
BO equals CO.
Therefore AO equals CO
Therefore by SSS triangle AFO is congruent to triangle CFO
and it follows that angles AFO and CFO are congruent.
Since these two angles AFO and CFO form a straight line,
they are both right.
Therefore FO is a perpendicular bisector of AC
Conclusion: The perpendicular bisectors of a triangle meet at
a common point - QED
4
The Conclusion brings
closure.
On Geometric Proofs:
Base angles of an isosceles trapezoid are equal
Perpendicular bisectors of a triangle meet at a common point
Before demonstrating the above proofs, we should review what sort of geometric tools are
available to us. Not all of what is presented below will be used but it’s useful to know what’s in
the arsenal.
1.
Triangle Congruence Relations: There are four: side-angle-side or SAS (Prop I.4) sideside-side or SSS (Prop I.8), angle-side-angle or ASA and angle-angle-side or AAS (Prop 1.26).
Corresponding sides and angles of congruent triangles are equal.
2.
Constructions I: Using straight edge and compass it is possible to
Bisect an angle (Prop, I.9)
Bisect a finite line segment (Prop I.10)
Construct a perpendicular line from any point on a given line (Prop I.11) or any point not
on a given line (Prop I.12)
3.
Vertical Angles (Prop I.15). If two lines cross the opposite (vertical) angles are equal.
That is 1 = 2
2
1
4.
Parallel Lines: If two parallel lines are crossed by a third, the alternate angles are equal,
1 = 2 , the exterior angle equal to the interior and opposite angle on the same side, 1 = 3 ,
and the two interior angles on the same side equal two right angles, 2 = 4 (Prop I.29).
1
4
2
3
In addition if two lines are crossed by a third such that the alternate angles are equal 1 = 2 or
the exterior angle is equal to the interior and opposite angle on the same side 1 = 3 or the two
interior angles on the same side equal two right angles 2 = 4 , then the lines are parallel (Prop
I.27 and I.28)
5.
Parallel Line Construction: Given a line and point not on the line it is possible to
construct a second line through the point that is parallel to the first line. (Prop I.31)
6.
Isosceles triangle result: The base angles of an isosceles triangle are equal (Prop I.5). If
the base angles of a triangle are equal then the triangle is isosceles (Prop I.6)
1
Theorem: The base angles of an isosceles trapezoid are equal
Introduction: A famous theorem of geometry proves the result
that the base angles of an isosceles triangle are equal. A similar
result holds for an isosceles trapezoid; that is in a trapezoid where
the lengths of the two non-parallel sides are equal (and the
parallel sides are of different length) the base angles are equal
Given an isosceles trapezoid ABCD where AD and BC are
parallel and AB equals CD, the base angles BAD and CDA are
equal. Assume BC is the shorter of the two parallel lines.
C
B
1
5
A
6 4
State what is being
proved
Background/
Introduction
Proposition (to be
proved)
3
2
7
D
E
Proof: Begin by constructing a line BE parallel to CD. Construct
the line CE.
Because diagonal CE intersects parallel lines BC and AD angle
BCE (angle 1) equals angle CED (angle 2).
Proof/Argument
Likewise because CE intersects parallel lines BE and CD angle
ECD (angle 3) equals angle CEB (angle 4).
Therefore since CE is common to both triangles ∆BEC and
∆DCE , triangles ∆BEC and ∆DCE are congruent by ASA.
Therefore BE equals CD and since CD equals AB, ∆ABE is
isosceles so angle BAE (angle 5) equals angle BEA (angle 6)
Since AD intersects parallel lines BE and CD angle BEA (angle 6)
equals angle CDA (angle 7).
Since angle BAD (angle 5) also equals angle BEA (angle 6),
angle BAD (angle 5) equals angle CDA (angle 7) and the result
follows.
Conclusion: Therefore the base angles of an isosceles trapezoid
are equal.
Conclusion/Summary
QED.
Corollary: Angle ABC equals angle DCB.
2
The purpose in writing “good mathematics” is to clearly and effectively communicate a
mathematical argument. I generally look for 4 things: clarity, correctness, conciseness,
completeness.
Clarity – This is the “biggie”: obviously any argument presented should be clear and easy to
understand. This is not always easy to do since mathematics can be difficult. Making a difficult
argument clear often presents a challenge; imagination is required. Presenting a good example or
examining a simple case often helps.
Correctness – obviously the logic of the argument can have no holes or flaws; this is the dealbreaker!
Conciseness – Wordiness of argument dulls the reader. An argument presented in 50 words is
easier to understand than an argument in 500 words
Completeness – Did you leave anything out? Did you tie off all the loose ends? Did you forget
something? Leave out an important statement necessary to make the whole argument work?
Completeness is the counter-balance to conciseness and effective writing has the right mix of
both.
A good argument should flow naturally – inexorably – to its conclusion.
The details of how you accomplish the above vary according to the assignment; there is no
universal template when presenting a mathematical argument (or proof). However I have found
that a good mathematical argument has four parts:
a statement (proposition/theorem) of what is to be proved or shown or argued,
an introduction/background which introduces the reader to what is to come,
the “proof” itself (the longest part of the write up) where all the work is done and finally
a short summary/conclusion which brings the whole write up to closure.
As an example the above is a (rather over done but effective) proof that the base angles of an
isosceles triangle are equal. Not all proofs can or need to be done this way but it illustrates one
way to structure a mathematical proof that is clear, correct, concise and complete.
We continue with a second proof that the perpendicular bisectors of a triangle meet at a common
point.
3
Theorem: The perpendicular bisectors of a triangle meet at a
common point.
Introduction: It’s not difficult to see that any two
perpendicular bisectors meet at a common point. We’ll show
that a line from the 3rd side to that common point is also a
perpendicular bisector
State what is being proved
A good introduction sets
up the reader for what is to
follow
Proof: Given triangle ABD
B
D
E
O
A
C
F
Let points D, E, and F be the bisectors of AB, BC, and AC
respectively.
Note the step by step
structure of the proof –
each statement naturally
following from the
previous. A good proof
should naturally flow
Construct the perpendicular bisectors DG and EH that meet at
point O.
By SAS triangle BDO is congruent to triangle ADO so AO
equals BO.
Again by SAS triangle BEO is congruent to triangle CEO so
BO equals CO.
Therefore AO equals CO
Therefore by SSS triangle AFO is congruent to triangle CFO
and it follows that angles AFO and CFO are congruent.
Since these two angles AFO and CFO form a straight line,
they are both right.
Therefore FO is a perpendicular bisector of AC
Conclusion: The perpendicular bisectors of a triangle meet at
a common point - QED
4
The Conclusion brings
closure.
On Geometric Proofs:
Base angles of an isosceles trapezoid are equal
Perpendicular bisectors of a triangle meet at a common point
Before demonstrating the above proofs, we should review what sort of geometric tools are
available to us. Not all of what is presented below will be used but it’s useful to know what’s in
the arsenal.
1.
Triangle Congruence Relations: There are four: side-angle-side or SAS (Prop I.4) sideside-side or SSS (Prop I.8), angle-side-angle or ASA and angle-angle-side or AAS (Prop 1.26).
Corresponding sides and angles of congruent triangles are equal.
2.
Constructions I: Using straight edge and compass it is possible to
Bisect an angle (Prop, I.9)
Bisect a finite line segment (Prop I.10)
Construct a perpendicular line from any point on a given line (Prop I.11) or any point not
on a given line (Prop I.12)
3.
Vertical Angles (Prop I.15). If two lines cross the opposite (vertical) angles are equal.
That is 1 = 2
2
1
4.
Parallel Lines: If two parallel lines are crossed by a third, the alternate angles are equal,
1 = 2 , the exterior angle equal to the interior and opposite angle on the same side, 1 = 3 ,
and the two interior angles on the same side equal two right angles, 2 = 4 (Prop I.29).
1
4
2
3
In addition if two lines are crossed by a third such that the alternate angles are equal 1 = 2 or
the exterior angle is equal to the interior and opposite angle on the same side 1 = 3 or the two
interior angles on the same side equal two right angles 2 = 4 , then the lines are parallel (Prop
I.27 and I.28)
5.
Parallel Line Construction: Given a line and point not on the line it is possible to
construct a second line through the point that is parallel to the first line. (Prop I.31)
6.
Isosceles triangle result: The base angles of an isosceles triangle are equal (Prop I.5). If
the base angles of a triangle are equal then the triangle is isosceles (Prop I.6)
1
Theorem: The base angles of an isosceles trapezoid are equal
Introduction: A famous theorem of geometry proves the result
that the base angles of an isosceles triangle are equal. A similar
result holds for an isosceles trapezoid; that is in a trapezoid where
the lengths of the two non-parallel sides are equal (and the
parallel sides are of different length) the base angles are equal
Given an isosceles trapezoid ABCD where AD and BC are
parallel and AB equals CD, the base angles BAD and CDA are
equal. Assume BC is the shorter of the two parallel lines.
C
B
1
5
A
6 4
State what is being
proved
Background/
Introduction
Proposition (to be
proved)
3
2
7
D
E
Proof: Begin by constructing a line BE parallel to CD. Construct
the line CE.
Because diagonal CE intersects parallel lines BC and AD angle
BCE (angle 1) equals angle CED (angle 2).
Proof/Argument
Likewise because CE intersects parallel lines BE and CD angle
ECD (angle 3) equals angle CEB (angle 4).
Therefore since CE is common to both triangles ∆BEC and
∆DCE , triangles ∆BEC and ∆DCE are congruent by ASA.
Therefore BE equals CD and since CD equals AB, ∆ABE is
isosceles so angle BAE (angle 5) equals angle BEA (angle 6)
Since AD intersects parallel lines BE and CD angle BEA (angle 6)
equals angle CDA (angle 7).
Since angle BAD (angle 5) also equals angle BEA (angle 6),
angle BAD (angle 5) equals angle CDA (angle 7) and the result
follows.
Conclusion: Therefore the base angles of an isosceles trapezoid
are equal.
Conclusion/Summary
QED.
Corollary: Angle ABC equals angle DCB.
2
The purpose in writing “good mathematics” is to clearly and effectively communicate a
mathematical argument. I generally look for 4 things: clarity, correctness, conciseness,
completeness.
Clarity – This is the “biggie”: obviously any argument presented should be clear and easy to
understand. This is not always easy to do since mathematics can be difficult. Making a difficult
argument clear often presents a challenge; imagination is required. Presenting a good example or
examining a simple case often helps.
Correctness – obviously the logic of the argument can have no holes or flaws; this is the dealbreaker!
Conciseness – Wordiness of argument dulls the reader. An argument presented in 50 words is
easier to understand than an argument in 500 words
Completeness – Did you leave anything out? Did you tie off all the loose ends? Did you forget
something? Leave out an important statement necessary to make the whole argument work?
Completeness is the counter-balance to conciseness and effective writing has the right mix of
both.
A good argument should flow naturally – inexorably – to its conclusion.
The details of how you accomplish the above vary according to the assignment; there is no
universal template when presenting a mathematical argument (or proof). However I have found
that a good mathematical argument has four parts:
a statement (proposition/theorem) of what is to be proved or shown or argued,
an introduction/background which introduces the reader to what is to come,
the “proof” itself (the longest part of the write up) where all the work is done and finally
a short summary/conclusion which brings the whole write up to closure.
As an example the above is a (rather over done but effective) proof that the base angles of an
isosceles triangle are equal. Not all proofs can or need to be done this way but it illustrates one
way to structure a mathematical proof that is clear, correct, concise and complete.
We continue with a second proof that the perpendicular bisectors of a triangle meet at a common
point.
3
Theorem: The perpendicular bisectors of a triangle meet at a
common point.
Introduction: It’s not difficult to see that any two
perpendicular bisectors meet at a common point. We’ll show
that a line from the 3rd side to that common point is also a
perpendicular bisector
State what is being proved
A good introduction sets
up the reader for what is to
follow
Proof: Given triangle ABD
B
D
E
O
A
C
F
Let points D, E, and F be the bisectors of AB, BC, and AC
respectively.
Note the step by step
structure of the proof –
each statement naturally
following from the
previous. A good proof
should naturally flow
Construct the perpendicular bisectors DG and EH that meet at
point O.
By SAS triangle BDO is congruent to triangle ADO so AO
equals BO.
Again by SAS triangle BEO is congruent to triangle CEO so
BO equals CO.
Therefore AO equals CO
Therefore by SSS triangle AFO is congruent to triangle CFO
and it follows that angles AFO and CFO are congruent.
Since these two angles AFO and CFO form a straight line,
they are both right.
Therefore FO is a perpendicular bisector of AC
Conclusion: The perpendicular bisectors of a triangle meet at
a common point - QED
4
The Conclusion brings
closure.
On Geometric Proofs:
Base angles of an isosceles trapezoid are equal
Perpendicular bisectors of a triangle meet at a common point
Before demonstrating the above proofs, we should review what sort of geometric tools are
available to us. Not all of what is presented below will be used but it’s useful to know what’s in
the arsenal.
1.
Triangle Congruence Relations: There are four: side-angle-side or SAS (Prop I.4) sideside-side or SSS (Prop I.8), angle-side-angle or ASA and angle-angle-side or AAS (Prop 1.26).
Corresponding sides and angles of congruent triangles are equal.
2.
Constructions I: Using straight edge and compass it is possible to
Bisect an angle (Prop, I.9)
Bisect a finite line segment (Prop I.10)
Construct a perpendicular line from any point on a given line (Prop I.11) or any point not
on a given line (Prop I.12)
3.
Vertical Angles (Prop I.15). If two lines cross the opposite (vertical) angles are equal.
That is 1 = 2
2
1
4.
Parallel Lines: If two parallel lines are crossed by a third, the alternate angles are equal,
1 = 2 , the exterior angle equal to the interior and opposite angle on the same side, 1 = 3 ,
and the two interior angles on the same side equal two right angles, 2 = 4 (Prop I.29).
1
4
2
3
In addition if two lines are crossed by a third such that the alternate angles are equal 1 = 2 or
the exterior angle is equal to the interior and opposite angle on the same side 1 = 3 or the two
interior angles on the same side equal two right angles 2 = 4 , then the lines are parallel (Prop
I.27 and I.28)
5.
Parallel Line Construction: Given a line and point not on the line it is possible to
construct a second line through the point that is parallel to the first line. (Prop I.31)
6.
Isosceles triangle result: The base angles of an isosceles triangle are equal (Prop I.5). If
the base angles of a triangle are equal then the triangle is isosceles (Prop I.6)
1
Theorem: The base angles of an isosceles trapezoid are equal
Introduction: A famous theorem of geometry proves the result
that the base angles of an isosceles triangle are equal. A similar
result holds for an isosceles trapezoid; that is in a trapezoid where
the lengths of the two non-parallel sides are equal (and the
parallel sides are of different length) the base angles are equal
Given an isosceles trapezoid ABCD where AD and BC are
parallel and AB equals CD, the base angles BAD and CDA are
equal. Assume BC is the shorter of the two parallel lines.
C
B
1
5
A
6 4
State what is being
proved
Background/
Introduction
Proposition (to be
proved)
3
2
7
D
E
Proof: Begin by constructing a line BE parallel to CD. Construct
the line CE.
Because diagonal CE intersects parallel lines BC and AD angle
BCE (angle 1) equals angle CED (angle 2).
Proof/Argument
Likewise because CE intersects parallel lines BE and CD angle
ECD (angle 3) equals angle CEB (angle 4).
Therefore since CE is common to both triangles ∆BEC and
∆DCE , triangles ∆BEC and ∆DCE are congruent by ASA.
Therefore BE equals CD and since CD equals AB, ∆ABE is
isosceles so angle BAE (angle 5) equals angle BEA (angle 6)
Since AD intersects parallel lines BE and CD angle BEA (angle 6)
equals angle CDA (angle 7).
Since angle BAD (angle 5) also equals angle BEA (angle 6),
angle BAD (angle 5) equals angle CDA (angle 7) and the result
follows.
Conclusion: Therefore the base angles of an isosceles trapezoid
are equal.
Conclusion/Summary
QED.
Corollary: Angle ABC equals angle DCB.
2
The purpose in writing “good mathematics” is to clearly and effectively communicate a
mathematical argument. I generally look for 4 things: clarity, correctness, conciseness,
completeness.
Clarity – This is the “biggie”: obviously any argument presented should be clear and easy to
understand. This is not always easy to do since mathematics can be difficult. Making a difficult
argument clear often presents a challenge; imagination is required. Presenting a good example or
examining a simple case often helps.
Correctness – obviously the logic of the argument can have no holes or flaws; this is the dealbreaker!
Conciseness – Wordiness of argument dulls the reader. An argument presented in 50 words is
easier to understand than an argument in 500 words
Completeness – Did you leave anything out? Did you tie off all the loose ends? Did you forget
something? Leave out an important statement necessary to make the whole argument work?
Completeness is the counter-balance to conciseness and effective writing has the right mix of
both.
A good argument should flow naturally – inexorably – to its conclusion.
The details of how you accomplish the above vary according to the assignment; there is no
universal template when presenting a mathematical argument (or proof). However I have found
that a good mathematical argument has four parts:
a statement (proposition/theorem) of what is to be proved or shown or argued,
an introduction/background which introduces the reader to what is to come,
the “proof” itself (the longest part of the write up) where all the work is done and finally
a short summary/conclusion which brings the whole write up to closure.
As an example the above is a (rather over done but effective) proof that the base angles of an
isosceles triangle are equal. Not all proofs can or need to be done this way but it illustrates one
way to structure a mathematical proof that is clear, correct, concise and complete.
We continue with a second proof that the perpendicular bisectors of a triangle meet at a common
point.
3
Theorem: The perpendicular bisectors of a triangle meet at a
common point.
Introduction: It’s not difficult to see that any two
perpendicular bisectors meet at a common point. We’ll show
that a line from the 3rd side to that common point is also a
perpendicular bisector
State what is being proved
A good introduction sets
up the reader for what is to
follow
Proof: Given triangle ABD
B
D
E
O
A
C
F
Let points D, E, and F be the bisectors of AB, BC, and AC
respectively.
Note the step by step
structure of the proof –
each statement naturally
following from the
previous. A good proof
should naturally flow
Construct the perpendicular bisectors DG and EH that meet at
point O.
By SAS triangle BDO is congruent to triangle ADO so AO
equals BO.
Again by SAS triangle BEO is congruent to triangle CEO so
BO equals CO.
Therefore AO equals CO
Therefore by SSS triangle AFO is congruent to triangle CFO
and it follows that angles AFO and CFO are congruent.
Since these two angles AFO and CFO form a straight line,
they are both right.
Therefore FO is a perpendicular bisector of AC
Conclusion: The perpendicular bisectors of a triangle meet at
a common point - QED
4
The Conclusion brings
closure.
On Geometric Proofs:
Base angles of an isosceles trapezoid are equal
Perpendicular bisectors of a triangle meet at a common point
Before demonstrating the above proofs, we should review what sort of geometric tools are
available to us. Not all of what is presented below will be used but it’s useful to know what’s in
the arsenal.
1.
Triangle Congruence Relations: There are four: side-angle-side or SAS (Prop I.4) sideside-side or SSS (Prop I.8), angle-side-angle or ASA and angle-angle-side or AAS (Prop 1.26).
Corresponding sides and angles of congruent triangles are equal.
2.
Constructions I: Using straight edge and compass it is possible to
Bisect an angle (Prop, I.9)
Bisect a finite line segment (Prop I.10)
Construct a perpendicular line from any point on a given line (Prop I.11) or any point not
on a given line (Prop I.12)
3.
Vertical Angles (Prop I.15). If two lines cross the opposite (vertical) angles are equal.
That is 1 = 2
2
1
4.
Parallel Lines: If two parallel lines are crossed by a third, the alternate angles are equal,
1 = 2 , the exterior angle equal to the interior and opposite angle on the same side, 1 = 3 ,
and the two interior angles on the same side equal two right angles, 2 = 4 (Prop I.29).
1
4
2
3
In addition if two lines are crossed by a third such that the alternate angles are equal 1 = 2 or
the exterior angle is equal to the interior and opposite angle on the same side 1 = 3 or the two
interior angles on the same side equal two right angles 2 = 4 , then the lines are parallel (Prop
I.27 and I.28)
5.
Parallel Line Construction: Given a line and point not on the line it is possible to
construct a second line through the point that is parallel to the first line. (Prop I.31)
6.
Isosceles triangle result: The base angles of an isosceles triangle are equal (Prop I.5). If
the base angles of a triangle are equal then the triangle is isosceles (Prop I.6)
1
Theorem: The base angles of an isosceles trapezoid are equal
Introduction: A famous theorem of geometry proves the result
that the base angles of an isosceles triangle are equal. A similar
result holds for an isosceles trapezoid; that is in a trapezoid where
the lengths of the two non-parallel sides are equal (and the
parallel sides are of different length) the base angles are equal
Given an isosceles trapezoid ABCD where AD and BC are
parallel and AB equals CD, the base angles BAD and CDA are
equal. Assume BC is the shorter of the two parallel lines.
C
B
1
5
A
6 4
State what is being
proved
Background/
Introduction
Proposition (to be
proved)
3
2
7
D
E
Proof: Begin by constructing a line BE parallel to CD. Construct
the line CE.
Because diagonal CE intersects parallel lines BC and AD angle
BCE (angle 1) equals angle CED (angle 2).
Proof/Argument
Likewise because CE intersects parallel lines BE and CD angle
ECD (angle 3) equals angle CEB (angle 4).
Therefore since CE is common to both triangles ∆BEC and
∆DCE , triangles ∆BEC and ∆DCE are congruent by ASA.
Therefore BE equals CD and since CD equals AB, ∆ABE is
isosceles so angle BAE (angle 5) equals angle BEA (angle 6)
Since AD intersects parallel lines BE and CD angle BEA (angle 6)
equals angle CDA (angle 7).
Since angle BAD (angle 5) also equals angle BEA (angle 6),
angle BAD (angle 5) equals angle CDA (angle 7) and the result
follows.
Conclusion: Therefore the base angles of an isosceles trapezoid
are equal.
Conclusion/Summary
QED.
Corollary: Angle ABC equals angle DCB.
2
The purpose in writing “good mathematics” is to clearly and effectively communicate a
mathematical argument. I generally look for 4 things: clarity, correctness, conciseness,
completeness.
Clarity – This is the “biggie”: obviously any argument presented should be clear and easy to
understand. This is not always easy to do since mathematics can be difficult. Making a difficult
argument clear often presents a challenge; imagination is required. Presenting a good example or
examining a simple case often helps.
Correctness – obviously the logic of the argument can have no holes or flaws; this is the dealbreaker!
Conciseness – Wordiness of argument dulls the reader. An argument presented in 50 words is
easier to understand than an argument in 500 words
Completeness – Did you leave anything out? Did you tie off all the loose ends? Did you forget
something? Leave out an important statement necessary to make the whole argument work?
Completeness is the counter-balance to conciseness and effective writing has the right mix of
both.
A good argument should flow naturally – inexorably – to its conclusion.
The details of how you accomplish the above vary according to the assignment; there is no
universal template when presenting a mathematical argument (or proof). However I have found
that a good mathematical argument has four parts:
a statement (proposition/theorem) of what is to be proved or shown or argued,
an introduction/background which introduces the reader to what is to come,
the “proof” itself (the longest part of the write up) where all the work is done and finally
a short summary/conclusion which brings the whole write up to closure.
As an example the above is a (rather over done but effective) proof that the base angles of an
isosceles triangle are equal. Not all proofs can or need to be done this way but it illustrates one
way to structure a mathematical proof that is clear, correct, concise and complete.
We continue with a second proof that the perpendicular bisectors of a triangle meet at a common
point.
3
Theorem: The perpendicular bisectors of a triangle meet at a
common point.
Introduction: It’s not difficult to see that any two
perpendicular bisectors meet at a common point. We’ll show
that a line from the 3rd side to that common point is also a
perpendicular bisector
State what is being proved
A good introduction sets
up the reader for what is to
follow
Proof: Given triangle ABD
B
D
E
O
A
C
F
Let points D, E, and F be the bisectors of AB, BC, and AC
respectively.
Note the step by step
structure of the proof –
each statement naturally
following from the
previous. A good proof
should naturally flow
Construct the perpendicular bisectors DG and EH that meet at
point O.
By SAS triangle BDO is congruent to triangle ADO so AO
equals BO.
Again by SAS triangle BEO is congruent to triangle CEO so
BO equals CO.
Therefore AO equals CO
Therefore by SSS triangle AFO is congruent to triangle CFO
and it follows that angles AFO and CFO are congruent.
Since these two angles AFO and CFO form a straight line,
they are both right.
Therefore FO is a perpendicular bisector of AC
Conclusion: The perpendicular bisectors of a triangle meet at
a common point - QED
4
The Conclusion brings
closure.
On Geometric Proofs:
Base angles of an isosceles trapezoid are equal
Perpendicular bisectors of a triangle meet at a common point
Before demonstrating the above proofs, we should review what sort of geometric tools are
available to us. Not all of what is presented below will be used but it’s useful to know what’s in
the arsenal.
1.
Triangle Congruence Relations: There are four: side-angle-side or SAS (Prop I.4) sideside-side or SSS (Prop I.8), angle-side-angle or ASA and angle-angle-side or AAS (Prop 1.26).
Corresponding sides and angles of congruent triangles are equal.
2.
Constructions I: Using straight edge and compass it is possible to
Bisect an angle (Prop, I.9)
Bisect a finite line segment (Prop I.10)
Construct a perpendicular line from any point on a given line (Prop I.11) or any point not
on a given line (Prop I.12)
3.
Vertical Angles (Prop I.15). If two lines cross the opposite (vertical) angles are equal.
That is 1 = 2
2
1
4.
Parallel Lines: If two parallel lines are crossed by a third, the alternate angles are equal,
1 = 2 , the exterior angle equal to the interior and opposite angle on the same side, 1 = 3 ,
and the two interior angles on the same side equal two right angles, 2 = 4 (Prop I.29).
1
4
2
3
In addition if two lines are crossed by a third such that the alternate angles are equal 1 = 2 or
the exterior angle is equal to the interior and opposite angle on the same side 1 = 3 or the two
interior angles on the same side equal two right angles 2 = 4 , then the lines are parallel (Prop
I.27 and I.28)
5.
Parallel Line Construction: Given a line and point not on the line it is possible to
construct a second line through the point that is parallel to the first line. (Prop I.31)
6.
Isosceles triangle result: The base angles of an isosceles triangle are equal (Prop I.5). If
the base angles of a triangle are equal then the triangle is isosceles (Prop I.6)
1
Theorem: The base angles of an isosceles trapezoid are equal
Introduction: A famous theorem of geometry proves the result
that the base angles of an isosceles triangle are equal. A similar
result holds for an isosceles trapezoid; that is in a trapezoid where
the lengths of the two non-parallel sides are equal (and the
parallel sides are of different length) the base angles are equal
Given an isosceles trapezoid ABCD where AD and BC are
parallel and AB equals CD, the base angles BAD and CDA are
equal. Assume BC is the shorter of the two parallel lines.
C
B
1
5
A
6 4
State what is being
proved
Background/
Introduction
Proposition (to be
proved)
3
2
7
D
E
Proof: Begin by constructing a line BE parallel to CD. Construct
the line CE.
Because diagonal CE intersects parallel lines BC and AD angle
BCE (angle 1) equals angle CED (angle 2).
Proof/Argument
Likewise because CE intersects parallel lines BE and CD angle
ECD (angle 3) equals angle CEB (angle 4).
Therefore since CE is common to both triangles ∆BEC and
∆DCE , triangles ∆BEC and ∆DCE are congruent by ASA.
Therefore BE equals CD and since CD equals AB, ∆ABE is
isosceles so angle BAE (angle 5) equals angle BEA (angle 6)
Since AD intersects parallel lines BE and CD angle BEA (angle 6)
equals angle CDA (angle 7).
Since angle BAD (angle 5) also equals angle BEA (angle 6),
angle BAD (angle 5) equals angle CDA (angle 7) and the result
follows.
Conclusion: Therefore the base angles of an isosceles trapezoid
are equal.
Conclusion/Summary
QED.
Corollary: Angle ABC equals angle DCB.
2
The purpose in writing “good mathematics” is to clearly and effectively communicate a
mathematical argument. I generally look for 4 things: clarity, correctness, conciseness,
completeness.
Clarity – This is the “biggie”: obviously any argument presented should be clear and easy to
understand. This is not always easy to do since mathematics can be difficult. Making a difficult
argument clear often presents a challenge; imagination is required. Presenting a good example or
examining a simple case often helps.
Correctness – obviously the logic of the argument can have no holes or flaws; this is the dealbreaker!
Conciseness – Wordiness of argument dulls the reader. An argument presented in 50 words is
easier to understand than an argument in 500 words
Completeness – Did you leave anything out? Did you tie off all the loose ends? Did you forget
something? Leave out an important statement necessary to make the whole argument work?
Completeness is the counter-balance to conciseness and effective writing has the right mix of
both.
A good argument should flow naturally – inexorably – to its conclusion.
The details of how you accomplish the above vary according to the assignment; there is no
universal template when presenting a mathematical argument (or proof). However I have found
that a good mathematical argument has four parts:
a statement (proposition/theorem) of what is to be proved or shown or argued,
an introduction/background which introduces the reader to what is to come,
the “proof” itself (the longest part of the write up) where all the work is done and finally
a short summary/conclusion which brings the whole write up to closure.
As an example the above is a (rather over done but effective) proof that the base angles of an
isosceles triangle are equal. Not all proofs can or need to be done this way but it illustrates one
way to structure a mathematical proof that is clear, correct, concise and complete.
We continue with a second proof that the perpendicular bisectors of a triangle meet at a common
point.
3
Theorem: The perpendicular bisectors of a triangle meet at a
common point.
Introduction: It’s not difficult to see that any two
perpendicular bisectors meet at a common point. We’ll show
that a line from the 3rd side to that common point is also a
perpendicular bisector
State what is being proved
A good introduction sets
up the reader for what is to
follow
Proof: Given triangle ABD
B
D
E
O
A
C
F
Let points D, E, and F be the bisectors of AB, BC, and AC
respectively.
Note the step by step
structure of the proof –
each statement naturally
following from the
previous. A good proof
should naturally flow
Construct the perpendicular bisectors DG and EH that meet at
point O.
By SAS triangle BDO is congruent to triangle ADO so AO
equals BO.
Again by SAS triangle BEO is congruent to triangle CEO so
BO equals CO.
Therefore AO equals CO
Therefore by SSS triangle AFO is congruent to triangle CFO
and it follows that angles AFO and CFO are congruent.
Since these two angles AFO and CFO form a straight line,
they are both right.
Therefore FO is a perpendicular bisector of AC
Conclusion: The perpendicular bisectors of a triangle meet at
a common point - QED
4
The Conclusion brings
closure.
On Geometric Proofs:
Base angles of an isosceles trapezoid are equal
Perpendicular bisectors of a triangle meet at a common point
Before demonstrating the above proofs, we should review what sort of geometric tools are
available to us. Not all of what is presented below will be used but it’s useful to know what’s in
the arsenal.
1.
Triangle Congruence Relations: There are four: side-angle-side or SAS (Prop I.4) sideside-side or SSS (Prop I.8), angle-side-angle or ASA and angle-angle-side or AAS (Prop 1.26).
Corresponding sides and angles of congruent triangles are equal.
2.
Constructions I: Using straight edge and compass it is possible to
Bisect an angle (Prop, I.9)
Bisect a finite line segment (Prop I.10)
Construct a perpendicular line from any point on a given line (Prop I.11) or any point not
on a given line (Prop I.12)
3.
Vertical Angles (Prop I.15). If two lines cross the opposite (vertical) angles are equal.
That is 1 = 2
2
1
4.
Parallel Lines: If two parallel lines are crossed by a third, the alternate angles are equal,
1 = 2 , the exterior angle equal to the interior and opposite angle on the same side, 1 = 3 ,
and the two interior angles on the same side equal two right angles, 2 = 4 (Prop I.29).
1
4
2
3
In addition if two lines are crossed by a third such that the alternate angles are equal 1 = 2 or
the exterior angle is equal to the interior and opposite angle on the same side 1 = 3 or the two
interior angles on the same side equal two right angles 2 = 4 , then the lines are parallel (Prop
I.27 and I.28)
5.
Parallel Line Construction: Given a line and point not on the line it is possible to
construct a second line through the point that is parallel to the first line. (Prop I.31)
6.
Isosceles triangle result: The base angles of an isosceles triangle are equal (Prop I.5). If
the base angles of a triangle are equal then the triangle is isosceles (Prop I.6)
1
Theorem: The base angles of an isosceles trapezoid are equal
Introduction: A famous theorem of geometry proves the result
that the base angles of an isosceles triangle are equal. A similar
result holds for an isosceles trapezoid; that is in a trapezoid where
the lengths of the two non-parallel sides are equal (and the
parallel sides are of different length) the base angles are equal
Given an isosceles trapezoid ABCD where AD and BC are
parallel and AB equals CD, the base angles BAD and CDA are
equal. Assume BC is the shorter of the two parallel lines.
C
B
1
5
A
6 4
State what is being
proved
Background/
Introduction
Proposition (to be
proved)
3
2
7
D
E
Proof: Begin by constructing a line BE parallel to CD. Construct
the line CE.
Because diagonal CE intersects parallel lines BC and AD angle
BCE (angle 1) equals angle CED (angle 2).
Proof/Argument
Likewise because CE intersects parallel lines BE and CD angle
ECD (angle 3) equals angle CEB (angle 4).
Therefore since CE is common to both triangles ∆BEC and
∆DCE , triangles ∆BEC and ∆DCE are congruent by ASA.
Therefore BE equals CD and since CD equals AB, ∆ABE is
isosceles so angle BAE (angle 5) equals angle BEA (angle 6)
Since AD intersects parallel lines BE and CD angle BEA (angle 6)
equals angle CDA (angle 7).
Since angle BAD (angle 5) also equals angle BEA (angle 6),
angle BAD (angle 5) equals angle CDA (angle 7) and the result
follows.
Conclusion: Therefore the base angles of an isosceles trapezoid
are equal.
Conclusion/Summary
QED.
Corollary: Angle ABC equals angle DCB.
2
The purpose in writing “good mathematics” is to clearly and effectively communicate a
mathematical argument. I generally look for 4 things: clarity, correctness, conciseness,
completeness.
Clarity – This is the “biggie”: obviously any argument presented should be clear and easy to
understand. This is not always easy to do since mathematics can be difficult. Making a difficult
argument clear often presents a challenge; imagination is required. Presenting a good example or
examining a simple case often helps.
Correctness – obviously the logic of the argument can have no holes or flaws; this is the dealbreaker!
Conciseness – Wordiness of argument dulls the reader. An argument presented in 50 words is
easier to understand than an argument in 500 words
Completeness – Did you leave anything out? Did you tie off all the loose ends? Did you forget
something? Leave out an important statement necessary to make the whole argument work?
Completeness is the counter-balance to conciseness and effective writing has the right mix of
both.
A good argument should flow naturally – inexorably – to its conclusion.
The details of how you accomplish the above vary according to the assignment; there is no
universal template when presenting a mathematical argument (or proof). However I have found
that a good mathematical argument has four parts:
a statement (proposition/theorem) of what is to be proved or shown or argued,
an introduction/background which introduces the reader to what is to come,
the “proof” itself (the longest part of the write up) where all the work is done and finally
a short summary/conclusion which brings the whole write up to closure.
As an example the above is a (rather over done but effective) proof that the base angles of an
isosceles triangle are equal. Not all proofs can or need to be done this way but it illustrates one
way to structure a mathematical proof that is clear, correct, concise and complete.
We continue with a second proof that the perpendicular bisectors of a triangle meet at a common
point.
3
Theorem: The perpendicular bisectors of a triangle meet at a
common point.
Introduction: It’s not difficult to see that any two
perpendicular bisectors meet at a common point. We’ll show
that a line from the 3rd side to that common point is also a
perpendicular bisector
State what is being proved
A good introduction sets
up the reader for what is to
follow
Proof: Given triangle ABD
B
D
E
O
A
C
F
Let points D, E, and F be the bisectors of AB, BC, and AC
respectively.
Note the step by step
structure of the proof –
each statement naturally
following from the
previous. A good proof
should naturally flow
Construct the perpendicular bisectors DG and EH that meet at
point O.
By SAS triangle BDO is congruent to triangle ADO so AO
equals BO.
Again by SAS triangle BEO is congruent to triangle CEO so
BO equals CO.
Therefore AO equals CO
Therefore by SSS triangle AFO is congruent to triangle CFO
and it follows that angles AFO and CFO are congruent.
Since these two angles AFO and CFO form a straight line,
they are both right.
Therefore FO is a perpendicular bisector of AC
Conclusion: The perpendicular bisectors of a triangle meet at
a common point - QED
4
The Conclusion brings
closure.
On Geometric Proofs:
Base angles of an isosceles trapezoid are equal
Perpendicular bisectors of a triangle meet at a common point
Before demonstrating the above proofs, we should review what sort of geometric tools are
available to us. Not all of what is presented below will be used but it’s useful to know what’s in
the arsenal.
1.
Triangle Congruence Relations: There are four: side-angle-side or SAS (Prop I.4) sideside-side or SSS (Prop I.8), angle-side-angle or ASA and angle-angle-side or AAS (Prop 1.26).
Corresponding sides and angles of congruent triangles are equal.
2.
Constructions I: Using straight edge and compass it is possible to
Bisect an angle (Prop, I.9)
Bisect a finite line segment (Prop I.10)
Construct a perpendicular line from any point on a given line (Prop I.11) or any point not
on a given line (Prop I.12)
3.
Vertical Angles (Prop I.15). If two lines cross the opposite (vertical) angles are equal.
That is 1 = 2
2
1
4.
Parallel Lines: If two parallel lines are crossed by a third, the alternate angles are equal,
1 = 2 , the exterior angle equal to the interior and opposite angle on the same side, 1 = 3 ,
and the two interior angles on the same side equal two right angles, 2 = 4 (Prop I.29).
1
4
2
3
In addition if two lines are crossed by a third such that the alternate angles are equal 1 = 2 or
the exterior angle is equal to the interior and opposite angle on the same side 1 = 3 or the two
interior angles on the same side equal two right angles 2 = 4 , then the lines are parallel (Prop
I.27 and I.28)
5.
Parallel Line Construction: Given a line and point not on the line it is possible to
construct a second line through the point that is parallel to the first line. (Prop I.31)
6.
Isosceles triangle result: The base angles of an isosceles triangle are equal (Prop I.5). If
the base angles of a triangle are equal then the triangle is isosceles (Prop I.6)
1
Theorem: The base angles of an isosceles trapezoid are equal
Introduction: A famous theorem of geometry proves the result
that the base angles of an isosceles triangle are equal. A similar
result holds for an isosceles trapezoid; that is in a trapezoid where
the lengths of the two non-parallel sides are equal (and the
parallel sides are of different length) the base angles are equal
Given an isosceles trapezoid ABCD where AD and BC are
parallel and AB equals CD, the base angles BAD and CDA are
equal. Assume BC is the shorter of the two parallel lines.
C
B
1
5
A
6 4
State what is being
proved
Background/
Introduction
Proposition (to be
proved)
3
2
7
D
E
Proof: Begin by constructing a line BE parallel to CD. Construct
the line CE.
Because diagonal CE intersects parallel lines BC and AD angle
BCE (angle 1) equals angle CED (angle 2).
Proof/Argument
Likewise because CE intersects parallel lines BE and CD angle
ECD (angle 3) equals angle CEB (angle 4).
Therefore since CE is common to both triangles ∆BEC and
∆DCE , triangles ∆BEC and ∆DCE are congruent by ASA.
Therefore BE equals CD and since CD equals AB, ∆ABE is
isosceles so angle BAE (angle 5) equals angle BEA (angle 6)
Since AD intersects parallel lines BE and CD angle BEA (angle 6)
equals angle CDA (angle 7).
Since angle BAD (angle 5) also equals angle BEA (angle 6),
angle BAD (angle 5) equals angle CDA (angle 7) and the result
follows.
Conclusion: Therefore the base angles of an isosceles trapezoid
are equal.
Conclusion/Summary
QED.
Corollary: Angle ABC equals angle DCB.
2
The purpose in writing “good mathematics” is to clearly and effectively communicate a
mathematical argument. I generally look for 4 things: clarity, correctness, conciseness,
completeness.
Clarity – This is the “biggie”: obviously any argument presented should be clear and easy to
understand. This is not always easy to do since mathematics can be difficult. Making a difficult
argument clear often presents a challenge; imagination is required. Presenting a good example or
examining a simple case often helps.
Correctness – obviously the logic of the argument can have no holes or flaws; this is the dealbreaker!
Conciseness – Wordiness of argument dulls the reader. An argument presented in 50 words is
easier to understand than an argument in 500 words
Completeness – Did you leave anything out? Did you tie off all the loose ends? Did you forget
something? Leave out an important statement necessary to make the whole argument work?
Completeness is the counter-balance to conciseness and effective writing has the right mix of
both.
A good argument should flow naturally – inexorably – to its conclusion.
The details of how you accomplish the above vary according to the assignment; there is no
universal template when presenting a mathematical argument (or proof). However I have found
that a good mathematical argument has four parts:
a statement (proposition/theorem) of what is to be proved or shown or argued,
an introduction/background which introduces the reader to what is to come,
the “proof” itself (the longest part of the write up) where all the work is done and finally
a short summary/conclusion which brings the whole write up to closure.
As an example the above is a (rather over done but effective) proof that the base angles of an
isosceles triangle are equal. Not all proofs can or need to be done this way but it illustrates one
way to structure a mathematical proof that is clear, correct, concise and complete.
We continue with a second proof that the perpendicular bisectors of a triangle meet at a common
point.
3
Theorem: The perpendicular bisectors of a triangle meet at a
common point.
Introduction: It’s not difficult to see that any two
perpendicular bisectors meet at a common point. We’ll show
that a line from the 3rd side to that common point is also a
perpendicular bisector
State what is being proved
A good introduction sets
up the reader for what is to
follow
Proof: Given triangle ABD
B
D
E
O
A
C
F
Let points D, E, and F be the bisectors of AB, BC, and AC
respectively.
Note the step by step
structure of the proof –
each statement naturally
following from the
previous. A good proof
should naturally flow
Construct the perpendicular bisectors DG and EH that meet at
point O.
By SAS triangle BDO is congruent to triangle ADO so AO
equals BO.
Again by SAS triangle BEO is congruent to triangle CEO so
BO equals CO.
Therefore AO equals CO
Therefore by SSS triangle AFO is congruent to triangle CFO
and it follows that angles AFO and CFO are congruent.
Since these two angles AFO and CFO form a straight line,
they are both right.
Therefore FO is a perpendicular bisector of AC
Conclusion: The perpendicular bisectors of a triangle meet at
a common point - QED
4
The Conclusion brings
closure.
On Geometric Proofs:
Base angles of an isosceles trapezoid are equal
Perpendicular bisectors of a triangle meet at a common point
Before demonstrating the above proofs, we should review what sort of geometric tools are
available to us. Not all of what is presented below will be used but it’s useful to know what’s in
the arsenal.
1.
Triangle Congruence Relations: There are four: side-angle-side or SAS (Prop I.4) sideside-side or SSS (Prop I.8), angle-side-angle or ASA and angle-angle-side or AAS (Prop 1.26).
Corresponding sides and angles of congruent triangles are equal.
2.
Constructions I: Using straight edge and compass it is possible to
Bisect an angle (Prop, I.9)
Bisect a finite line segment (Prop I.10)
Construct a perpendicular line from any point on a given line (Prop I.11) or any point not
on a given line (Prop I.12)
3.
Vertical Angles (Prop I.15). If two lines cross the opposite (vertical) angles are equal.
That is 1 = 2
2
1
4.
Parallel Lines: If two parallel lines are crossed by a third, the alternate angles are equal,
1 = 2 , the exterior angle equal to the interior and opposite angle on the same side, 1 = 3 ,
and the two interior angles on the same side equal two right angles, 2 = 4 (Prop I.29).
1
4
2
3
In addition if two lines are crossed by a third such that the alternate angles are equal 1 = 2 or
the exterior angle is equal to the interior and opposite angle on the same side 1 = 3 or the two
interior angles on the same side equal two right angles 2 = 4 , then the lines are parallel (Prop
I.27 and I.28)
5.
Parallel Line Construction: Given a line and point not on the line it is possible to
construct a second line through the point that is parallel to the first line. (Prop I.31)
6.
Isosceles triangle result: The base angles of an isosceles triangle are equal (Prop I.5). If
the base angles of a triangle are equal then the triangle is isosceles (Prop I.6)
1
Theorem: The base angles of an isosceles trapezoid are equal
Introduction: A famous theorem of geometry proves the result
that the base angles of an isosceles triangle are equal. A similar
result holds for an isosceles trapezoid; that is in a trapezoid where
the lengths of the two non-parallel sides are equal (and the
parallel sides are of different length) the base angles are equal
Given an isosceles trapezoid ABCD where AD and BC are
parallel and AB equals CD, the base angles BAD and CDA are
equal. Assume BC is the shorter of the two parallel lines.
C
B
1
5
A
6 4
State what is being
proved
Background/
Introduction
Proposition (to be
proved)
3
2
7
D
E
Proof: Begin by constructing a line BE parallel to CD. Construct
the line CE.
Because diagonal CE intersects parallel lines BC and AD angle
BCE (angle 1) equals angle CED (angle 2).
Proof/Argument
Likewise because CE intersects parallel lines BE and CD angle
ECD (angle 3) equals angle CEB (angle 4).
Therefore since CE is common to both triangles ∆BEC and
∆DCE , triangles ∆BEC and ∆DCE are congruent by ASA.
Therefore BE equals CD and since CD equals AB, ∆ABE is
isosceles so angle BAE (angle 5) equals angle BEA (angle 6)
Since AD intersects parallel lines BE and CD angle BEA (angle 6)
equals angle CDA (angle 7).
Since angle BAD (angle 5) also equals angle BEA (angle 6),
angle BAD (angle 5) equals angle CDA (angle 7) and the result
follows.
Conclusion: Therefore the base angles of an isosceles trapezoid
are equal.
Conclusion/Summary
QED.
Corollary: Angle ABC equals angle DCB.
2
The purpose in writing “good mathematics” is to clearly and effectively communicate a
mathematical argument. I generally look for 4 things: clarity, correctness, conciseness,
completeness.
Clarity – This is the “biggie”: obviously any argument presented should be clear and easy to
understand. This is not always easy to do since mathematics can be difficult. Making a difficult
argument clear often presents a challenge; imagination is required. Presenting a good example or
examining a simple case often helps.
Correctness – obviously the logic of the argument can have no holes or flaws; this is the dealbreaker!
Conciseness – Wordiness of argument dulls the reader. An argument presented in 50 words is
easier to understand than an argument in 500 words
Completeness – Did you leave anything out? Did you tie off all the loose ends? Did you forget
something? Leave out an important statement necessary to make the whole argument work?
Completeness is the counter-balance to conciseness and effective writing has the right mix of
both.
A good argument should flow naturally – inexorably – to its conclusion.
The details of how you accomplish the above vary according to the assignment; there is no
universal template when presenting a mathematical argument (or proof). However I have found
that a good mathematical argument has four parts:
a statement (proposition/theorem) of what is to be proved or shown or argued,
an introduction/background which introduces the reader to what is to come,
the “proof” itself (the longest part of the write up) where all the work is done and finally
a short summary/conclusion which brings the whole write up to closure.
As an example the above is a (rather over done but effective) proof that the base angles of an
isosceles triangle are equal. Not all proofs can or need to be done this way but it illustrates one
way to structure a mathematical proof that is clear, correct, concise and complete.
We continue with a second proof that the perpendicular bisectors of a triangle meet at a common
point.
3
Theorem: The perpendicular bisectors of a triangle meet at a
common point.
Introduction: It’s not difficult to see that any two
perpendicular bisectors meet at a common point. We’ll show
that a line from the 3rd side to that common point is also a
perpendicular bisector
State what is being proved
A good introduction sets
up the reader for what is to
follow
Proof: Given triangle ABD
B
D
E
O
A
C
F
Let points D, E, and F be the bisectors of AB, BC, and AC
respectively.
Note the step by step
structure of the proof –
each statement naturally
following from the
previous. A good proof
should naturally flow
Construct the perpendicular bisectors DG and EH that meet at
point O.
By SAS triangle BDO is congruent to triangle ADO so AO
equals BO.
Again by SAS triangle BEO is congruent to triangle CEO so
BO equals CO.
Therefore AO equals CO
Therefore by SSS triangle AFO is congruent to triangle CFO
and it follows that angles AFO and CFO are congruent.
Since these two angles AFO and CFO form a straight line,
they are both right.
Therefore FO is a perpendicular bisector of AC
Conclusion: The perpendicular bisectors of a triangle meet at
a common point - QED
4
The Conclusion brings
closure.
On Geometric Proofs:
Base angles of an isosceles trapezoid are equal
Perpendicular bisectors of a triangle meet at a common point
Before demonstrating the above proofs, we should review what sort of geometric tools are
available to us. Not all of what is presented below will be used but it’s useful to know what’s in
the arsenal.
1.
Triangle Congruence Relations: There are four: side-angle-side or SAS (Prop I.4) sideside-side or SSS (Prop I.8), angle-side-angle or ASA and angle-angle-side or AAS (Prop 1.26).
Corresponding sides and angles of congruent triangles are equal.
2.
Constructions I: Using straight edge and compass it is possible to
Bisect an angle (Prop, I.9)
Bisect a finite line segment (Prop I.10)
Construct a perpendicular line from any point on a given line (Prop I.11) or any point not
on a given line (Prop I.12)
3.
Vertical Angles (Prop I.15). If two lines cross the opposite (vertical) angles are equal.
That is 1 = 2
2
1
4.
Parallel Lines: If two parallel lines are crossed by a third, the alternate angles are equal,
1 = 2 , the exterior angle equal to the interior and opposite angle on the same side, 1 = 3 ,
and the two interior angles on the same side equal two right angles, 2 = 4 (Prop I.29).
1
4
2
3
In addition if two lines are crossed by a third such that the alternate angles are equal 1 = 2 or
the exterior angle is equal to the interior and opposite angle on the same side 1 = 3 or the two
interior angles on the same side equal two right angles 2 = 4 , then the lines are parallel (Prop
I.27 and I.28)
5.
Parallel Line Construction: Given a line and point not on the line it is possible to
construct a second line through the point that is parallel to the first line. (Prop I.31)
6.
Isosceles triangle result: The base angles of an isosceles triangle are equal (Prop I.5). If
the base angles of a triangle are equal then the triangle is isosceles (Prop I.6)
1
Theorem: The base angles of an isosceles trapezoid are equal
Introduction: A famous theorem of geometry proves the result
that the base angles of an isosceles triangle are equal. A similar
result holds for an isosceles trapezoid; that is in a trapezoid where
the lengths of the two non-parallel sides are equal (and the
parallel sides are of different length) the base angles are equal
Given an isosceles trapezoid ABCD where AD and BC are
parallel and AB equals CD, the base angles BAD and CDA are
equal. Assume BC is the shorter of the two parallel lines.
C
B
1
5
A
6 4
State what is being
proved
Background/
Introduction
Proposition (to be
proved)
3
2
7
D
E
Proof: Begin by constructing a line BE parallel to CD. Construct
the line CE.
Because diagonal CE intersects parallel lines BC and AD angle
BCE (angle 1) equals angle CED (angle 2).
Proof/Argument
Likewise because CE intersects parallel lines BE and CD angle
ECD (angle 3) equals angle CEB (angle 4).
Therefore since CE is common to both triangles ∆BEC and
∆DCE , triangles ∆BEC and ∆DCE are congruent by ASA.
Therefore BE equals CD and since CD equals AB, ∆ABE is
isosceles so angle BAE (angle 5) equals angle BEA (angle 6)
Since AD intersects parallel lines BE and CD angle BEA (angle 6)
equals angle CDA (angle 7).
Since angle BAD (angle 5) also equals angle BEA (angle 6),
angle BAD (angle 5) equals angle CDA (angle 7) and the result
follows.
Conclusion: Therefore the base angles of an isosceles trapezoid
are equal.
Conclusion/Summary
QED.
Corollary: Angle ABC equals angle DCB.
2
The purpose in writing “good mathematics” is to clearly and effectively communicate a
mathematical argument. I generally look for 4 things: clarity, correctness, conciseness,
completeness.
Clarity – This is the “biggie”: obviously any argument presented should be clear and easy to
understand. This is not always easy to do since mathematics can be difficult. Making a difficult
argument clear often presents a challenge; imagination is required. Presenting a good example or
examining a simple case often helps.
Correctness – obviously the logic of the argument can have no holes or flaws; this is the dealbreaker!
Conciseness – Wordiness of argument dulls the reader. An argument presented in 50 words is
easier to understand than an argument in 500 words
Completeness – Did you leave anything out? Did you tie off all the loose ends? Did you forget
something? Leave out an important statement necessary to make the whole argument work?
Completeness is the counter-balance to conciseness and effective writing has the right mix of
both.
A good argument should flow naturally – inexorably – to its conclusion.
The details of how you accomplish the above vary according to the assignment; there is no
universal template when presenting a mathematical argument (or proof). However I have found
that a good mathematical argument has four parts:
a statement (proposition/theorem) of what is to be proved or shown or argued,
an introduction/background which introduces the reader to what is to come,
the “proof” itself (the longest part of the write up) where all the work is done and finally
a short summary/conclusion which brings the whole write up to closure.
As an example the above is a (rather over done but effective) proof that the base angles of an
isosceles triangle are equal. Not all proofs can or need to be done this way but it illustrates one
way to structure a mathematical proof that is clear, correct, concise and complete.
We continue with a second proof that the perpendicular bisectors of a triangle meet at a common
point.
3
Theorem: The perpendicular bisectors of a triangle meet at a
common point.
Introduction: It’s not difficult to see that any two
perpendicular bisectors meet at a common point. We’ll show
that a line from the 3rd side to that common point is also a
perpendicular bisector
State what is being proved
A good introduction sets
up the reader for what is to
follow
Proof: Given triangle ABD
B
D
E
O
A
C
F
Let points D, E, and F be the bisectors of AB, BC, and AC
respectively.
Note the step by step
structure of the proof –
each statement naturally
following from the
previous. A good proof
should naturally flow
Construct the perpendicular bisectors DG and EH that meet at
point O.
By SAS triangle BDO is congruent to triangle ADO so AO
equals BO.
Again by SAS triangle BEO is congruent to triangle CEO so
BO equals CO.
Therefore AO equals CO
Therefore by SSS triangle AFO is congruent to triangle CFO
and it follows that angles AFO and CFO are congruent.
Since these two angles AFO and CFO form a straight line,
they are both right.
Therefore FO is a perpendicular bisector of AC
Conclusion: The perpendicular bisectors of a triangle meet at
a common point - QED
4
The Conclusion brings
closure.
On Geometric Proofs:
Base angles of an isosceles trapezoid are equal
Perpendicular bisectors of a triangle meet at a common point
Before demonstrating the above proofs, we should review what sort of geometric tools are
available to us. Not all of what is presented below will be used but it’s useful to know what’s in
the arsenal.
1.
Triangle Congruence Relations: There are four: side-angle-side or SAS (Prop I.4) sideside-side or SSS (Prop I.8), angle-side-angle or ASA and angle-angle-side or AAS (Prop 1.26).
Corresponding sides and angles of congruent triangles are equal.
2.
Constructions I: Using straight edge and compass it is possible to
Bisect an angle (Prop, I.9)
Bisect a finite line segment (Prop I.10)
Construct a perpendicular line from any point on a given line (Prop I.11) or any point not
on a given line (Prop I.12)
3.
Vertical Angles (Prop I.15). If two lines cross the opposite (vertical) angles are equal.
That is 1 = 2
2
1
4.
Parallel Lines: If two parallel lines are crossed by a third, the alternate angles are equal,
1 = 2 , the exterior angle equal to the interior and opposite angle on the same side, 1 = 3 ,
and the two interior angles on the same side equal two right angles, 2 = 4 (Prop I.29).
1
4
2
3
In addition if two lines are crossed by a third such that the alternate angles are equal 1 = 2 or
the exterior angle is equal to the interior and opposite angle on the same side 1 = 3 or the two
interior angles on the same side equal two right angles 2 = 4 , then the lines are parallel (Prop
I.27 and I.28)
5.
Parallel Line Construction: Given a line and point not on the line it is possible to
construct a second line through the point that is parallel to the first line. (Prop I.31)
6.
Isosceles triangle result: The base angles of an isosceles triangle are equal (Prop I.5). If
the base angles of a triangle are equal then the triangle is isosceles (Prop I.6)
1
Theorem: The base angles of an isosceles trapezoid are equal
Introduction: A famous theorem of geometry proves the result
that the base angles of an isosceles triangle are equal. A similar
result holds for an isosceles trapezoid; that is in a trapezoid where
the lengths of the two non-parallel sides are equal (and the
parallel sides are of different length) the base angles are equal
Given an isosceles trapezoid ABCD where AD and BC are
parallel and AB equals CD, the base angles BAD and CDA are
equal. Assume BC is the shorter of the two parallel lines.
C
B
1
5
A
6 4
State what is being
proved
Background/
Introduction
Proposition (to be
proved)
3
2
7
D
E
Proof: Begin by constructing a line BE parallel to CD. Construct
the line CE.
Because diagonal CE intersects parallel lines BC and AD angle
BCE (angle 1) equals angle CED (angle 2).
Proof/Argument
Likewise because CE intersects parallel lines BE and CD angle
ECD (angle 3) equals angle CEB (angle 4).
Therefore since CE is common to both triangles ∆BEC and
∆DCE , triangles ∆BEC and ∆DCE are congruent by ASA.
Therefore BE equals CD and since CD equals AB, ∆ABE is
isosceles so angle BAE (angle 5) equals angle BEA (angle 6)
Since AD intersects parallel lines BE and CD angle BEA (angle 6)
equals angle CDA (angle 7).
Since angle BAD (angle 5) also equals angle BEA (angle 6),
angle BAD (angle 5) equals angle CDA (angle 7) and the result
follows.
Conclusion: Therefore the base angles of an isosceles trapezoid
are equal.
Conclusion/Summary
QED.
Corollary: Angle ABC equals angle DCB.
2
The purpose in writing “good mathematics” is to clearly and effectively communicate a
mathematical argument. I generally look for 4 things: clarity, correctness, conciseness,
completeness.
Clarity – This is the “biggie”: obviously any argument presented should be clear and easy to
understand. This is not always easy to do since mathematics can be difficult. Making a difficult
argument clear often presents a challenge; imagination is required. Presenting a good example or
examining a simple case often helps.
Correctness – obviously the logic of the argument can have no holes or flaws; this is the dealbreaker!
Conciseness – Wordiness of argument dulls the reader. An argument presented in 50 words is
easier to understand than an argument in 500 words
Completeness – Did you leave anything out? Did you tie off all the loose ends? Did you forget
something? Leave out an important statement necessary to make the whole argument work?
Completeness is the counter-balance to conciseness and effective writing has the right mix of
both.
A good argument should flow naturally – inexorably – to its conclusion.
The details of how you accomplish the above vary according to the assignment; there is no
universal template when presenting a mathematical argument (or proof). However I have found
that a good mathematical argument has four parts:
a statement (proposition/theorem) of what is to be proved or shown or argued,
an introduction/background which introduces the reader to what is to come,
the “proof” itself (the longest part of the write up) where all the work is done and finally
a short summary/conclusion which brings the whole write up to closure.
As an example the above is a (rather over done but effective) proof that the base angles of an
isosceles triangle are equal. Not all proofs can or need to be done this way but it illustrates one
way to structure a mathematical proof that is clear, correct, concise and complete.
We continue with a second proof that the perpendicular bisectors of a triangle meet at a common
point.
3
Theorem: The perpendicular bisectors of a triangle meet at a
common point.
Introduction: It’s not difficult to see that any two
perpendicular bisectors meet at a common point. We’ll show
that a line from the 3rd side to that common point is also a
perpendicular bisector
State what is being proved
A good introduction sets
up the reader for what is to
follow
Proof: Given triangle ABD
B
D
E
O
A
C
F
Let points D, E, and F be the bisectors of AB, BC, and AC
respectively.
Note the step by step
structure of the proof –
each statement naturally
following from the
previous. A good proof
should naturally flow
Construct the perpendicular bisectors DG and EH that meet at
point O.
By SAS triangle BDO is congruent to triangle ADO so AO
equals BO.
Again by SAS triangle BEO is congruent to triangle CEO so
BO equals CO.
Therefore AO equals CO
Therefore by SSS triangle AFO is congruent to triangle CFO
and it follows that angles AFO and CFO are congruent.
Since these two angles AFO and CFO form a straight line,
they are both right.
Therefore FO is a perpendicular bisector of AC
Conclusion: The perpendicular bisectors of a triangle meet at
a common point - QED
4
The Conclusion brings
closure.
On Geometric Proofs:
Base angles of an isosceles trapezoid are equal
Perpendicular bisectors of a triangle meet at a common point
Before demonstrating the above proofs, we should review what sort of geometric tools are
available to us. Not all of what is presented below will be used but it’s useful to know what’s in
the arsenal.
1.
Triangle Congruence Relations: There are four: side-angle-side or SAS (Prop I.4) sideside-side or SSS (Prop I.8), angle-side-angle or ASA and angle-angle-side or AAS (Prop 1.26).
Corresponding sides and angles of congruent triangles are equal.
2.
Constructions I: Using straight edge and compass it is possible to
Bisect an angle (Prop, I.9)
Bisect a finite line segment (Prop I.10)
Construct a perpendicular line from any point on a given line (Prop I.11) or any point not
on a given line (Prop I.12)
3.
Vertical Angles (Prop I.15). If two lines cross the opposite (vertical) angles are equal.
That is 1 = 2
2
1
4.
Parallel Lines: If two parallel lines are crossed by a third, the alternate angles are equal,
1 = 2 , the exterior angle equal to the interior and opposite angle on the same side, 1 = 3 ,
and the two interior angles on the same side equal two right angles, 2 = 4 (Prop I.29).
1
4
2
3
In addition if two lines are crossed by a third such that the alternate angles are equal 1 = 2 or
the exterior angle is equal to the interior and opposite angle on the same side 1 = 3 or the two
interior angles on the same side equal two right angles 2 = 4 , then the lines are parallel (Prop
I.27 and I.28)
5.
Parallel Line Construction: Given a line and point not on the line it is possible to
construct a second line through the point that is parallel to the first line. (Prop I.31)
6.
Isosceles triangle result: The base angles of an isosceles triangle are equal (Prop I.5). If
the base angles of a triangle are equal then the triangle is isosceles (Prop I.6)
1
Theorem: The base angles of an isosceles trapezoid are equal
Introduction: A famous theorem of geometry proves the result
that the base angles of an isosceles triangle are equal. A similar
result holds for an isosceles trapezoid; that is in a trapezoid where
the lengths of the two non-parallel sides are equal (and the
parallel sides are of different length) the base angles are equal
Given an isosceles trapezoid ABCD where AD and BC are
parallel and AB equals CD, the base angles BAD and CDA are
equal. Assume BC is the shorter of the two parallel lines.
C
B
1
5
A
6 4
State what is being
proved
Background/
Introduction
Proposition (to be
proved)
3
2
7
D
E
Proof: Begin by constructing a line BE parallel to CD. Construct
the line CE.
Because diagonal CE intersects parallel lines BC and AD angle
BCE (angle 1) equals angle CED (angle 2).
Proof/Argument
Likewise because CE intersects parallel lines BE and CD angle
ECD (angle 3) equals angle CEB (angle 4).
Therefore since CE is common to both triangles ∆BEC and
∆DCE , triangles ∆BEC and ∆DCE are congruent by ASA.
Therefore BE equals CD and since CD equals AB, ∆ABE is
isosceles so angle BAE (angle 5) equals angle BEA (angle 6)
Since AD intersects parallel lines BE and CD angle BEA (angle 6)
equals angle CDA (angle 7).
Since angle BAD (angle 5) also equals angle BEA (angle 6),
angle BAD (angle 5) equals angle CDA (angle 7) and the result
follows.
Conclusion: Therefore the base angles of an isosceles trapezoid
are equal.
Conclusion/Summary
QED.
Corollary: Angle ABC equals angle DCB.
2
The purpose in writing “good mathematics” is to clearly and effectively communicate a
mathematical argument. I generally look for 4 things: clarity, correctness, conciseness,
completeness.
Clarity – This is the “biggie”: obviously any argument presented should be clear and easy to
understand. This is not always easy to do since mathematics can be difficult. Making a difficult
argument clear often presents a challenge; imagination is required. Presenting a good example or
examining a simple case often helps.
Correctness – obviously the logic of the argument can have no holes or flaws; this is the dealbreaker!
Conciseness – Wordiness of argument dulls the reader. An argument presented in 50 words is
easier to understand than an argument in 500 words
Completeness – Did you leave anything out? Did you tie off all the loose ends? Did you forget
something? Leave out an important statement necessary to make the whole argument work?
Completeness is the counter-balance to conciseness and effective writing has the right mix of
both.
A good argument should flow naturally – inexorably – to its conclusion.
The details of how you accomplish the above vary according to the assignment; there is no
universal template when presenting a mathematical argument (or proof). However I have found
that a good mathematical argument has four parts:
a statement (proposition/theorem) of what is to be proved or shown or argued,
an introduction/background which introduces the reader to what is to come,
the “proof” itself (the longest part of the write up) where all the work is done and finally
a short summary/conclusion which brings the whole write up to closure.
As an example the above is a (rather over done but effective) proof that the base angles of an
isosceles triangle are equal. Not all proofs can or need to be done this way but it illustrates one
way to structure a mathematical proof that is clear, correct, concise and complete.
We continue with a second proof that the perpendicular bisectors of a triangle meet at a common
point.
3
Theorem: The perpendicular bisectors of a triangle meet at a
common point.
Introduction: It’s not difficult to see that any two
perpendicular bisectors meet at a common point. We’ll show
that a line from the 3rd side to that common point is also a
perpendicular bisector
State what is being proved
A good introduction sets
up the reader for what is to
follow
Proof: Given triangle ABD
B
D
E
O
A
C
F
Let points D, E, and F be the bisectors of AB, BC, and AC
respectively.
Note the step by step
structure of the proof –
each statement naturally
following from the
previous. A good proof
should naturally flow
Construct the perpendicular bisectors DG and EH that meet at
point O.
By SAS triangle BDO is congruent to triangle ADO so AO
equals BO.
Again by SAS triangle BEO is congruent to triangle CEO so
BO equals CO.
Therefore AO equals CO
Therefore by SSS triangle AFO is congruent to triangle CFO
and it follows that angles AFO and CFO are congruent.
Since these two angles AFO and CFO form a straight line,
they are both right.
Therefore FO is a perpendicular bisector of AC
Conclusion: The perpendicular bisectors of a triangle meet at
a common point - QED
4
The Conclusion brings
closure.