Download More on Neutral Geometry I (Including Section 3.3) ( "NIB" means

More on Neutral Geometry I (Including Section 3.3)
( "NIB" means "NOT IN BOOK" )
Theorem (NIB), The "The Adjacent Supplementary Angles" Theorem
(Converse of Postulate 14) :
If two adjacent angles are supplementary, then they form a linear pair.
Proof: Suppose that ∠ ABD and
−−−−→
ray BD ) are supplementary.
∠ DBC
are two adjacent angles (with common
F
(We need to show that they form a linear pair.)
C
B
A
By Postulate 4 (The Ruler Placement Postulate),
←−−−→
there is a point on line AB such that A – B – F .
D
Thus, since F and C are both in the half-plane side of
←−−−→
BD opposite the half-plane containing A,
←−−−→
F and C are on the same side of BD .
Now,
∠ ABD
and
∠ DBF
form a linear pair by definition of “linear pair”.
By Postulate 14, ∠ ABD and ∠ DBF are supplementary. By the supposition above,
∠ ABD and ∠ DBC are also supplementary. Therefore, ∠ DBC ≅ ∠ DBF ./
∴ m (∠ DBC ) = m (∠ DBF ) .
−−−−→
−−−−→
We see then that ray BC and BF are both rays in the same half-plane side of
←−−−→
−−−−→
line BD such that each ray combines with ray BD to produce angle with the
same measure r = 180° – m(∠ ABD) . By Postulate 12 (The Angle Construction
←−−−→
Postulate), there is one and only one ray in that same half-plane side of BD which
−−−−→
unions with BD to produce the angle with that particular angle measure, r =
−−−−→
m(∠ DBC ) = m(∠ DBF ) . By the uniqueness of this ray, we can conclude that BC =
−−−−→
BF
and so,
QED
∠ DBC = ∠ DBF
.
Therefore, ∠ ‫ ܥܤܦ ∠ ݀݊ܽ ܦܤܣ‬form a linear pair.
2
Section 3.3: Triangle Congruence Conditions
Theorem 3.3.1 (Angle-Side-Angle Congruence Condition):
If, in two triangles, the vertices of one triangle can be put into one-to-one correspondence with
the vertices of the other triangle such that:
Two angles and the included side of one triangle are congruent to the corresponding angles
and included side of the other triangle,
Then, the correspondence of triangle vertices is a congruence and the two triangles are
congruent.
Proof: Let ∆ ABC and ∆ DEF be two triangles and consider the correspondence of vertices
ABC DEF . Suppose that ∠ CAB ≅ ∠ FDE , and ∠ CBA ≅ ∠ FED , and AB ≅ DE .
C
F
A
B
D
E
[We need to show ∆ ABC ≅ ∆ DEF . ]
[ We do this by showing that AC ≅ DF , and then we apply Postulate 15 (SAS) to conclude that
∆ ABC ≅ ∆ DEF .]
Suppose, by way of contradiction, that AC and DF are not congruent . ∴ AC ≠ DF .
Without loss of generality, we assume that AC > DF .
[ Note: If AC < DF , then we rename the points A, B, C as D, E, F, and we rename the
points D, E, F as A, B, C , respectively. After these renamings, every statement made to this
point is true. In particular, AC > DF . ]
Since AC > DF, there exists a point T such that A–T–C and AT = DF , by the Ruler
Placement Postulate (Postulate 4). Note that ∠ TAB ≅ ∠ FDE since ∠ TAB = ∠ CAB .
C
F
T
A
B
D
E
Since m( ∠ ABC ) = m( ∠ DEF ) , we conclude by Postulate 13 that m( ∠ ABT ) < m( ∠ DEF ) .
In particular, m( ∠ ABT ) ≠ m( ∠ DEF ) . Since AT = DF , AT ≅ DF . Recall that
∠ TAB ≅ ∠ FDE and AB ≅ DE . Therefore, ∆ ABT ≅ ∆ DEF by SAS , and
∴ ∠ ABT ≅ ∠ DEF by CPCF and ∴ m( ∠ ABT ) = m( ∠ DEF ) , which contradicts that fact
that m( ∠ ABT ) < m( ∠ DEF ) . ∴ AC ≅ DF .
∴ ∆ ABC ≅ ∆ DEF by SAS . ∴ The ASA Triangle Congruence Condition is valid. Q E D
3
Theorem 3.3.2 , The "Congruent Base Angles" Theorem
(Converse of the "Isosceles Triangle Theorem"):
If two angles in a triangle are congruent, then the sides opposite these congruent angles are
congruent segments and the triangle is an Isosceles triangle.
Proof: The proof is a simple application of the A-S-A triangle congruence condition and is
left as an exercise.
Theorem 3.3.3 (Angle-Angle-Side Congruence Condition):
If, in two triangles, the vertices of one triangle can be put into one-to-one correspondence with
the vertices of the other triangle such that:
Two angles and the side opposite one of them in one triangle are congruent to the
corresponding angles and corresponding opposite side of the other triangle,
Then, the correspondence of triangle vertices is a congruence and the two triangles are
congruent.
Proof: Let ∆ ABC and ∆ DEF be two triangles and consider the correspondence of vertices
ABC DEF . Suppose that ∠ CAB ≅ ∠ FDE , and ∠ ABC ≅ ∠ DEF and AC ≅ DF .
C
F
A
D
B
E
[ We show that AB ≅ DE using a proof-by-contradiction. ]
Suppose, by way of contradiction, that AB and DE are not congruent . ∴ AB ≠ DE .
Without loss of generality, we assume that AB > DE . Since AB > DE, there exists a point
T such that A–T–B and AT = DE , by the Ruler Placement Postulate (Postulate 4). Thus,
AT ≅ DE .
C
A
F
T
B
D
E
Now, ∠ ATC is an exterior angle of ∆ CTB and ∠ TBC is an interior angle of that triangle
which is not adjacent to ∠ ATC .
∴ By Theorem 3.2.9 (The Exterior Angle Theorem),
m( ∠ ATC ) > m( ∠ TBC ) .
4
Note that ∠ CAT ≅ ∠ FDE since ∠ CAT = ∠ CAB , and recall that AC ≅ DF and AT ≅ DE .
∴ ∆ ATC ≅ ∆ DEF by Postulate 15 (SAS) . ∴ ∠ ATC ≅ ∠ DEF by CPCF and , therefore, m(
∠ ATC ) = m( ∠ DEF ) .
Also, m( ∠ DEF ) = m( ∠ ABC ) = m( ∠ TBC ) since ∠ TBC = ∠ ABC .
∴m( ∠ ATC ) = m( ∠ TBC ) , which contradicts the fact that m( ∠ ATC ) > m( ∠ TBC ) .
∴ AB ≅ DE .
Recall that ∠ CAB ≅ ∠ FDE and AC ≅ DF . ∴ ∆ ABC ≅ ∆ DEF by SAS .
∴ The AAS Triangle Congruence Condition is valid. Q E D
Theorem 3.3.4: SASAS Congruence for Quadrilaterals
Theorem 3.3.5, The "Larger Side opposite Larger Angle" Theorem:
For two sides of a triangle and the angles opposite them,
the angle opposite the larger side is the larger of the twos angles.
Theorem 3.3.7 – The Triangle Inequality:
A
B
C
The sum of the lengths of any two sides > the length of the third side.
For proving the Side-Side-Side Congruence Condition, we will need the Hinge Theorem which
is not proven here, but a proof of it can be found in the textbook.
Theorem 3.3.8 – The Hinge Theorem:
Given that two sides of one triangle are congruent to two corresponding sides of
another triangle, then, the greater of the two included angles has the longer of the two
corresponding third sides opposite these included angles.
F
C
The Shorter Side is Opposite
The Longer Side is Opposite
of the Smaller Angle.
A
B
of the Larger Angle.
D
E
5
Theorem 3.3.9 (Side-Side-Side Congruence Condition):
If, in two triangles, the vertices of one triangle can be put into one-to-one correspondence with
the vertices of the other triangle such that:
The three sides of one triangle are congruent to the corresponding sides of the other triangle,
Then, the correspondence of triangle vertices is a congruence and the two triangles are
congruent.
Proof: Let ∆ ABC and ∆ DEF be two triangles and consider the correspondence of vertices
ABC DEF . Suppose that AC ≅ DF and AB ≅ DE and BC ≅ EF .
F
C
A
B
D
E
[ We show that ∠ CAB ≅ ∠ FDE using a proof-by-contradiction. Then, we conclude that
∴ ∆ ABC ≅ ∆ DEF by SAS . ] Suppose, by way of contradiction, ∠ CAB and ∠ FDE are
not congruent. ∴m( ∠ CAB ) ≠ m( ∠ FDE ). Without loss of generality, we assume that m(
∠ CAB ) > m( ∠ FDE ) . Also, recall that AC ≅ DF and AB ≅ DE . By Theorem 3.3.8 (The
Hinge Theorem), BC > EF , which contradicts the fact that BC ≅ EF .
∴ ∠ CAB ≅ ∠ FDE . ∴ ∆ ABC ≅ ∆ DEF by Postulate 15 (SAS) .
∴ The SSS Triangle Congruence Condition is valid.
QED
Theorem (NIB) , The “Drop a Perpendicular” Theorem.
Given any line l and any point P, there exists a unique line through P which is perpendicular
to line l .
Proof: Consider line l and a point P.
We first assume that point P is not on line l . (See the figure.)
P
l
We must show that there exists a line through P which is perpendicular to line l.
6
←−→
Let A be any point on line l . If line PA is perpendicular to line l , then such a perpendicular line exists.
←−→
Suppose then that line PA is not perpendicular to line l.
Let B be a point on l so that ∠ BAP is an acute angle.
P
By the Angle Construction Postulate, there is a unique
−−−→
ray AF in the half-plane on the side of line l which is
opposite of the side containing P such that
l
D
H
A
B
m (∠ BAF ) = m (∠ BAP ) .
−−−→
C
By the Ruler Postulate, there is a point C on ray AC such
F
that AC = AP. Points P and C are on different half-plane
sides of l ; therefore, line l intersects the segment PC at
some point D with C – D – P.
By construction, ∠ DAC = ∠ DAF ≅ ∠ DAP . By construction,
AC ≅ AP . Also, AD ≅ AD . Therefore, ∆ADC ≅ ∆ADP by SAS.
∴ ∠ ADC ≅ ∠ ADP by CPCF.
∴ ∠ ADC and ∠ ADP are right angles by the Congruent Supplementary Angles Theorem. Therefore, line
←−→
PC is perpendicular to line l . Therefore, there exists at least one line through point P which is perpendicular
to l.
To prove that there is only one such line through P perpendicular to line l , suppose (by way of contradiction)
that there were a second line through P and
perpendicular to line l intersecting line l at point E .
Without loss of generality, we can assume that
P
A – D – E. (See the figure.)
Then, ∠ ADP is an exterior angle of ∆DEP .
By the Exterior Angle Theorem,
m( ∠ ADP ) > m( ∠ DEP ) since ∠ DEP is a nonadjacent interior angle of ∆DEP . But they are both
right angles: m( ∠ ADP ) = m( ∠ DEP ) = 90°, a
contradiction. Thus, when point P is not on line l ,
there is a unique line through P which is perpendicular to line l .
Line l
A
E
D
C
F
We next assume that point P is on line l .
Since the point P is on line l , then, by the Angle Construction
Postulate, in each half plane bounded by line l , there is one and only
one ray from point P forming an angle of measure 90°. By the
"Adjacent Supplementary Angles" Theorem (Converse of SMSG
Postulate 14) , the rays in the two half-planes are opposite rays that
form the unique line through P perpendicular to line l.
QED
P
l