Download September 17, 2014

Parallelogram
Exactly 1 pair of
parallel sides
2 pairs of parallel
sides
Exactly 1 pair of
congruent sides
2 pairs of
congruent sides
All sides
congruent
Perpendicular
diagonals
Congruent
diagonals
Diagonals bisect
angles
Diagonals bisect
each other
Opposite angles
congruent
Four right angles
Base angles
congruent
Rhombus
Rectangle
Square
Trapezoid
Isosceles
Trapezoid
Parallelogram
Rhombus
Rectangle
Square
Trapezoid
Isosceles
Trapezoid
Exactly 1 pair of
What
it take to prove that a quadrilateral is a rhombus?
parallel does
sides
2 pairsWe
of parallel
know a rhombus has perpendicular diagonals.
sides
Does
this mean that if we have a quadrilateral with perpendicular
Exactly 1 pair of
congruent sides
diagonals, it must be a rhombus?
2 pairs of
congruent sides
All sides
congruent
Perpendicular
diagonals
Congruent
diagonals
Diagonals bisect
angles
Diagonals bisect
each other
Opposite angles
congruent
Four right angles
Base angles
congruent
Ways to prove that a quadrilateral is a parallelogram
Ways to prove that a quadrilateral is a parallelogram
Show both pairs of opposite sides parallel
Ways to prove that a quadrilateral is a parallelogram
Show both pairs of opposite sides parallel
Show one pair of opposite sides parallel and congruent
C
D
A
B
Ways to prove that a quadrilateral is a parallelogram
Show both pairs of opposite sides parallel
Show one pair of opposite sides parallel and congruent
Show both pairs of opposite sides congruent
Show both pairs of opposite angles congruent
Show that the diagonals bisect each other
Ways to prove that a quadrilateral is a parallelogram
Show both pairs of opposite sides parallel
Show one pair of opposite sides parallel and congruent
Show both pairs of opposite sides congruent
Show both pairs of opposite angles congruent
Show that the diagonals bisect each other
Ways to prove that a quadrilateral is a rhombus
parallelogram and two adjacent sides congruent
Ways to prove that a quadrilateral is a parallelogram
Show both pairs of opposite sides parallel
Show one pair of opposite sides parallel and congruent
Show both pairs of opposite sides congruent
Show both pairs of opposite angles congruent
Show that the diagonals bisect each other
Ways to prove that a quadrilateral is a rhombus
parallelogram and two adjacent sides congruent
parallelogram and perpendicular diagonals
C
D
A
B
Ways to prove that a quadrilateral is a parallelogram
Show both pairs of opposite sides parallel
Show one pair of opposite sides parallel and congruent
Show both pairs of opposite sides congruent
Show both pairs of opposite angles congruent
Show that the diagonals bisect each other
Ways to prove that a quadrilateral is a rhombus
parallelogram and two adjacent sides congruent
parallelogram and perpendicular diagonals
parallelogram and diagonals bisect its angles
Ways to prove that a quadrilateral is a parallelogram
Show both pairs of opposite sides parallel
Show one pair of opposite sides parallel and congruent
Show both pairs of opposite sides congruent
Show both pairs of opposite angles congruent
Show that the diagonals bisect each other
Ways to prove that a quadrilateral is a rhombus
parallelogram and two adjacent sides congruent
parallelogram and perpendicular diagonals
parallelogram and diagonals bisect its angles
Ways to prove that a quadrilateral is a rectangle
parallelogram and one right angle
parallelogram and congruent diagonals
Ways to prove that a quadrilateral is an isosceles trapezoid
trapezoid and non-parallel sides congruent
trapezoid and one pair of congruent base angles
trapezoid and congruent diagonals
B
We know that an altitude of a triangle is a segment
from a vertex that is perpendicular to the opposite side
(AD in the diagram).
D
A
The median of a triangle is a segment from a vertex
to the midpoint of the opposite side.
A

B
C
M
AM is a median of triangle ABC.
C
A
MN is a midsegment
of triangle ABC
N
M
C
B
A
MN is the median
or midsegment of
trapezoid ABCD
M
D
B
N
C
The segment joining the midpoints of two sides of a triangle
(the midsegment) is parallel to the third side and half its length.
A
½x
M
N
x
C
B
The median of a trapezoid is parallel to the bases and is
equal in length to the average of the lengths of the bases.
A
M
D
b1
½ (b1 + b2)
B
N
b2
C
The segment joining the midpoints of two sides of a triangle
(the midsegment) is parallel to the third side and half its length.
A
M
B
N

C
P
The segment joining the midpoints of two sides of a triangle
(the midsegment) is parallel to the third side and half its length.
AM  PC, A  PCA (CPCTC)
( alternate interior angles)
MP  BC
A
M
B
N

C
P
The segment joining the midpoints of two sides of a triangle
(the midsegment) is parallel to the third side and half its length.
AM  PC, A  PCA (CPCTC)
( alternate interior angles)
MP  BC
A
M
B
N

C
P
The segment joining the midpoints of two sides of a triangle
(the midsegment) is parallel to the third side and half its length.
AM  PC, A  PCA (CPCTC)
( alternate interior angles)
MP  BC
A
M
B
N

C
P
The segment joining the midpoints of two sides of a triangle
(the midsegment) is parallel to the third side and half its length.
AM  PC, A  PCA (CPCTC)
( alternate interior angles)
MP  BC
A
M
B
N

C
P
The segment joining the midpoints of two sides of a triangle
(the midsegment) is parallel to the third side and half its length.
AM  PC, A  PCA (CPCTC)
( alternate interior angles)
MP  BC
A
M
B
N

C
P
The segment joining the midpoints of two sides of a triangle
(the midsegment) is parallel to the third side and half its length.
AM  PC, A  PCA (CPCTC)
( alternate interior angles)
MP  BC
A
M
B
N

C
P
The segment joining the midpoints of two sides of a triangle
(the midsegment) is parallel to the third side and half its length.
AM  PC, A  PCA (CPCTC)
( alternate interior angles)
MP  BC
A
M
B
N

C
P
The median of a trapezoid is parallel to the bases and is
equal in length to the average of the lengths of the bases.
A
M
D
B
N
C
The median of a trapezoid is parallel to the bases and is
equal in length to the average of the lengths of the bases.
Proof:
A
M
D
B
P

N
C
Because MP is the midsegment of ABD, MP is parallel to AB by the
previous theorem.
The median of a trapezoid is parallel to the bases and is
equal in length to the average of the lengths of the bases.
Proof:
A
M
D
B
P

N
C
Because MP is the midsegment of ABD, MP is parallel to AB by the
previous theorem. Since MP is part of MN, MN is parallel to AB.
The median of a trapezoid is parallel to the bases and is
equal in length to the average of the lengths of the bases.
Proof:
A
M
D
B
P

N
C
Because MP is the midsegment of ABD, MP is parallel to AB by the
previous theorem. Since MP is part of MN, MN is parallel to AB.
Similarly, PN is the midsegment of ADC, so that PN is parallel to DC.
Since PN is part of MN, MN is parallel to DC.
Therefore, median MN is parallel to bases AB and CD.
The median of a trapezoid is parallel to the bases and is
equal in length to the average of the lengths of the bases.
Proof:
A
M
D
B
P

N
C
Because MP is the midsegment of ABD, MP is parallel to AB by the
previous theorem. Since MP is part of MN, MN is parallel to AB.
Where is the flaw in this proof?
Similarly, PN is the midsegment of ADC, so that PN is parallel to DC.
Since PN is part of MN, MN is parallel to DC.
Therefore, median MN is parallel to bases AB and CD.
The median of a trapezoid is parallel to the bases and is
equal in length to the average of the lengths of the bases.
Proof:
A
M
D
B
N
C
First we prove the following lemma:
If two lines are parallel to the same line,
they are parallel to each other.
If m // l and n // l, prove m // n
l
The median of a trapezoid is parallel to the bases and is
equal in length to the average of the lengths of the bases.
Proof:
A
M
D
B
N
C
First we prove the following lemma:
If two lines are parallel to the same line,
they are parallel to each other.
If m // l and n // l, prove m // n
l
The median of a trapezoid is parallel to the bases and is
equal in length to the average of the lengths of the bases.
Proof:
A
M
D
B
N
C
First we prove the following lemma:
If two lines are parallel to the same line,
they are parallel to each other.
If m // l and n // l, prove m // n
l
The median of a trapezoid is parallel to the bases and is
equal in length to the average of the lengths of the bases.
Proof:
A
M
B
P

D
MP is parallel to AB.
It is a midsegment of ABD
PN is parallel to DC.
It is a midsegment of BDC
N
C
DC is parallel to AB.
Definition of trapezoid
Therefore, by the lemma,
MP is parallel to DC.
(they are both parallel to AB)
But Euclid’s Parallel Postulate says there can
only be one line through P parallel to DC.
Therefore,MN
PN and
MP must be
same
line,
i.e. DC.
MN.
Therefore,
is parallel
tothe
both
AB
and
The median of a trapezoid is parallel to the bases and is
equal in length to the average of the lengths of the bases.
Proof:
A
M
D
b1
½ b1P

B
½ b2
N
b2
MN = ½ b1 + ½ b2 = ½ (b1 + b2)
C
The segment joining the midpoints of two sides of a triangle
(the midsegment) is parallel to the third side and half its length.
A
½x
M
N
x
C
B
The median of a trapezoid is parallel to the bases and is
equal in length to the average of the lengths of the bases.
A
M
D
b1
½29(binches
1 + b2 )
B
N
42binches
2
C
The altitude of a parallelogram
C
B
The altitude of a trapezoid
D
A
D
A
C
B
The following construction problems should be worked on and solved
in groups of three people. Each group will be assessed three times—
once for each category. When your group is ready to be assessed,
notify me and indicate on which category the group wants to be
assessed. Your group will choose the problem in that category on
which to be assessed. The group member who will be asked to
solve the problem will be selected randomly. Whatever grade is
assigned to that student will be the grade for the group. Be sure each
group member can solve each problem correctly. The following
scoring guidelines will be used to assess each problem. The
maximum possible point total is 45 points.
Construction
0
Cannot do the construction correctly even with hints
3
Can do the construction correctly but only after hints are given
6
Can do the construction correctly without hints
The following construction problems should be worked on and solved
in groups of three people. Each group will be assessed three times—
once for each category. When your group is ready to be assessed,
notify me and indicate on which category the group wants to be
assessed. Your group will choose the problem in that category on
which to be assessed. The group member who will be asked to
solve the problem will be selected randomly. Whatever grade is
assigned to that student will be the grade for the group. Be sure each
group member can solve each problem correctly. The following
scoring guidelines will be used to assess each problem. The
maximum possible point total is 45 points.
Proof
0
Cannot do the proof correctly even with hints
3
Can do the proof correctly but only after hints are given
6
Can do the proof correctly without hints
Sample: Construct a rhombus given one side and an altitude
1. Construct a rhombus (not a square), given
a. one side and one angle (1 point)
b. one angle and a diagonal (2 points)
c. the altitude and one diagonal (3 points)
2. Construct a parallelogram (not a rhombus), given
a. one side, one angle, and one diagonal (1 point)
b. two adjacent sides and an altitude (2 points)
c. one angle, one side, and the altitude on that side
(3 points)
3. Construct an isosceles trapezoid, given
a. the diagonal, altitude, and one of the bases (1 point)
b. one base, the diagonal, and the angle included by them (2 points)
c. the bases and one angle (3 points)
Construct a rhombus given
one side and an altitude.
Construct a rhombus given
one side and an altitude.
Solution
1. Begin by copying one of the sides on the “Figures” sheet and label it AB.
Construct a rhombus given
one side and an altitude.
Solution
1. Begin by copying one of the sides on the “Figures” sheet and label it AB.
2. Choose a point P between A and B
Construct a rhombus given
one side and an altitude.
Solution
1. Begin by copying one of the sides on the “Figures” sheet and label it AB.
2. Choose a point P between A and B and construct a perpendicular to AB through P.
Construct a rhombus given
one side and an altitude.
Solution
1. Begin by copying one of the sides on the “Figures” sheet and label it AB.
2. Choose a point P between A and B and construct a perpendicular to AB through P.
3. Copy the altitude from the “figures” sheet so that one its endpoints is point P and
the other is on the perpendicular you constructed. Label the other endpoint Q.
Construct a rhombus given
one side and an altitude.
Solution
1. Begin by copying one of the sides on the “Figures” sheet and label it AB.
2. Choose a point P between A and B and construct a perpendicular to AB through P.
3. Copy the altitude from the “figures” sheet so that one its endpoints is point P and
the other is on the perpendicular you constructed. Label the other endpoint Q.
4. Using point A as center, make a circle with radius AB.
Construct a rhombus given
one side and an altitude.
Solution
1. Begin by copying one of the sides on the “Figures” sheet and label it AB.
2. Choose a point P between A and B and construct a perpendicular to AB through P.
3. Copy the altitude from the “figures” sheet so that one its endpoints is point P and
the other is on the perpendicular you constructed. Label the other endpoint Q.
4. Using point A as center, make a circle with radius AB.
5. Construct a line through point Q perpendicular to PQ (construction marks not
shown). Label the point where it intersects the circle point R.
Construct a rhombus given
one side and an altitude.
Solution
1. Begin by copying one of the sides on the “Figures” sheet and label it AB.
2. Choose a point P between A and B and construct a perpendicular to AB through P.
3. Copy the altitude from the “figures” sheet so that one its endpoints is point P and
the other is on the perpendicular you constructed. Label the other endpoint Q.
4. Using point A as center, make a circle with radius AB.
5. Construct a line through point Q perpendicular to PQ (construction marks not
shown). Label the point where it intersects the circle point R.
6. Construct radius AR.
Construct a rhombus given
one side and an altitude.
Solution
1. Begin by copying one of the sides on the “Figures” sheet and label it AB.
2. Choose a point P between A and B and construct a perpendicular to AB through P.
3. Copy the altitude from the “figures” sheet so that one its endpoints is point P and
the other is on the perpendicular you constructed. Label the other endpoint Q.
4. Using point A as center, make a circle with radius AB.
5. Construct a line through point Q perpendicular to PQ (construction marks not
shown). Label the point where it intersects the circle point R.
6. Construct radius AR.
7. Using point R as center, mark off a length equal to AB along ray QR. Call the
intersection point T.
Construct a rhombus given
one side and an altitude.
Conclusion: ARTB is a rhombus with
the required information.
Solution
1. Begin by copying one of the sides on the “Figures” sheet and label it AB.
2. Choose a point P between A and B and construct a perpendicular to AB through P.
3. Copy the altitude from the “figures” sheet so that one its endpoints is point P and
the other is on the perpendicular you constructed. Label the other endpoint Q.
4. Using point A as center, make a circle with radius AB.
5. Construct a line through point Q perpendicular to PQ (construction marks not
shown). Label the point where it intersects the circle point R.
6. Construct radius AR.
7. Using point R as center, mark off a length equal to AB along ray QR. Call the
intersection point T.
8. Construct BT.
Construct a rhombus given
one side and an altitude.
Conclusion: ARTB is a rhombus with
the required information.
Brief Proof:
Because AR, AB, and RT, are all radii of congruent
circles, they have the same length. Because RT
and AB are both perpendicular to PQ,
RT is parallel to AB. Therefore, ARTB a
parallelogram (one pair of opposite sides are both
congruent and parallel). It is now also a rhombus
because two adjacent sides are congruent (AR  AB).
The sides of the rhombus are all congruent to AB,
which was the given side. Since QT is parallel to AB,
the distance from QT to AB is always the same, and
that distance is equal to PQ, the given altitude.