Download Parallelogram Bisectors – Geometry Final Part B – Problem 7

Parallelogram Bisectors – Geometry Final Part B – Problem 7
By: Douglas A. Ruby
Class: Geometry
Date: 11/10/2002
Grades: 11/12
Problem 7: When the bisectors of two consecutive angles of a parallelogram intersect at a
common point on the remaining side, what can you conclude about each of the three triangles
formed? Prove each conclusion. Construct a supporting figure using Sketchpad.
Discussion:
Creating an arbitrary parallelogram using Sketchpad is relatively easy. One creates 3 arbitrary
points, constructs a vertex angle, and then constructs two more lines parallel to the first two
segments that pass through the two points at the end of each segment. However, arbitrary
parallelograms may produce bisectors that intersect inside or outside the parallelogram, rather
than on the remaining side. We see this in the next two constructions:
Experimentation using Sketchpad led to the discovery that to construct a parallelogram such that
the consecutive bisectors intersected on the remaining side, we had to use a specific type of
parallelogram. (Subject of Proof 1)
Page 1
Consecutive Bisectors of a Parallelogram
This is a parallelogram that has two sides equal to twice the length of the other two sides. In
order to construct a parallelogram whose sides are in a 2:1 ratio, we started with a segment (AD)
and then created a second segment at an arbitrary angle via rotation. We then dilated that
segment by a factor of 2 to create the full 2nd side of the parallelogram (in this case segment DC).
Next, we create the opposite two sides in the customary way already described (segments AB
and BC). Finally, we create the angle bisectors and find out that they intersect on the remaining
side. We see this in the next construction:
Using GSP, we explored the internal measurements of our construction in the next part of our
construction.
From these measurements and previous constructions, we observe that:
1. ADE and BCE are both isosceles triangles whose congruent sides are equal to the
smaller of the sides of the parallelogram.
2. ADE is an isosceles triangle whose base angles are each ½ of CDA.
3. BCE is an isosceles triangle whose base angles are each ½ of BCD.
4. DEC is always a right triangle (regardless of whether point E is on segment AB or not).
Page 2
Consecutive Bisectors of a Parallelogram
Proof #1 of the Bisectors of the Consecutive Angles of a Parallelogram
Theorem:
Given:
Prove:
When the consecutive angle bisectors of a parallelogram intersect on a point on
the remaining side of the parallelogram, that point divides that side into two
segments, exactly equal to the smaller of the two sides of the parallelogram.
Parallelogram ABCD
Point E on AB
DE bisects CDA, CE bisects
BCD
Point E bisects AB, AB = 2 × AD
(See Next Page)
Page 3
1.
2.
3.
4.
5.
6.
Statement
Parallelogram ABCD
Point E on AB
DE bisects CDA, CE bisects
AD || BC
AB || CD
ADE
CDE
7.
DCE
BCE
8.
AED
CDE
9.
AED
ADE
10. ADE is isosceles
11. BEC
DCE
12. BEC
BCE
13. BCE is isosceles
14. AD
AE
15. BC
BE
16. AD
BC
17. AD BE
18. AE BE
19. Point E bisects AB
20. AB = AE + BE
21. AB = AD + AD
22. AB = 2 × AD
Reason
BCD
Given
Given
Given
Opposite sides of a parallelogram are ||
Opposite sides of a parallelogram are ||
Angle bisector divides an angle into two
congruent angles
Angle bisector divides an angle into two
congruent angles
Alternate interior angles
Substitution from 6 and 8
A triangle with two congruent base angles is
isosceles
Alternate interior angles
Substitution from 7 and 11
A triangle with two congruent base angles is
isosceles
Sides opposite base angles of an isosceles
triangle are congruent
Sides opposite base angles of an isosceles
triangle are congruent
Opposite sides of a parallelogram are
congruent
Substitution from 15 and 16
Transitive property from 14 and 17
Segment bisector is a point that divides a
segment into two equal halves
Segment addition postulate
Substitution from 14, 17, and 18
Distributive property of multiplication
We have proven that when the consecutive angle bisectors of a parallelogram intersect on a point
on the remaining side of the parallelogram, that point divides that side into two segments, exactly
equal to the smaller of the two sides of the parallelogram. Put another way, the ratio of the
adjacent sides of the parallelogram must be 2:1.
With Proof #1, we also showed that:
1.
ADE and BCE are both isosceles triangles whose congruent sides are equal to the
smaller of the sides of the parallelogram.
Now, we want to prove the remaining three observations in Proof#2:
2. ADE is an isosceles triangle whose base angles are each ½ of CDA.
3. BCE is an isosceles triangle whose base angles are each ½ of BCD.
4. DEC is always a right triangle (regardless of whether point E is on segment AB or not).
Page 4
Consecutive Bisectors of a Parallelogram
Proof #2 of the Bisectors of the Consecutive Angles of a Parallelogram
Theorem:
When the consecutive angle bisectors of a parallelogram intersect on a point on
the remaining side of the parallelogram, one of the triangles formed is a right
triangle, while the other two are isosceles.
Given:
Parallelogram ABCD
Point E on AB
DE bisects CDA, CE bisects BCD
Point E bisects AB, AB = 2 × AD (from Proof #1)
ADE and BCE are both isosceles triangles (from Proof #1)
Prove:
m ADE = m AED = m CDA/2.
m BCE = m BEC = m BCD/2.
DEC is a right triangle
(See Next Page)
Page 5
1.
2.
3.
4.
5.
6.
7.
Statement
Parallelogram ABCD
Point E on AB
DE bisects CDA, CE bisects BCD
Point E bisects AB
AB = 2 × AD
ADE and BCE are isosceles
ADE
CDE
8.
CDE
AED
9.
ADE
AED
10. m ADE = m AED
11. m CDA = m ADE + m AED
12. m CDA = m ADE + m ADE
13. m CDA = 2 × m ADE
14. m CDA/2 = m ADE
15. m ADE = m AED = m CDA/2
16. DCE
BCE
17. BEC
DCE
18. BCE
BEC
19. m BCE = m BEC
20. m BCD = m DCE + m BCE
21. m BCD = m BCE + m BCE
22. m BCD = 2 × m BCE
23. m BCD/2 = m BCE
24. m BCE = m BEC = m BCD/2
25. m CDA + m BCD = 180o
26. (m CDA + m BCD)/2 = 180o/2 = 90o
27. m CDA/2 + m BCD/2 = 90o
28. m CDE = m CDA/2
29. m DCE = m BCD/2
30. m CDE + m DCE = 90o
31. m CDE + m DCE + m DEC = 180o
32. m DEC = 180o – (m CDE + m DCE)
33. m DEC= 180o – 90o = 90o
34. DEC is a right triangle
Final Part B – Problem 7 – Doug Ruby - Page 6
Reason
Given
Given
Given
From Proof #1
From Proof #1
From Proof #1
Angle bisector divides an angle into two
congruent angles
Alternate interior angles
Substitution from 7 and 8
Congruent angles have equal measure
Angle Addition Postulate
Substitution from 10 and 11
Distributive Property of Multiplication
Identity property of division
Substitution from 10 and 15
Angle bisector divides an angle into two
congruent angles
Alternate interior angles
Substitution from 16 and 17
Congruent angles have equal measure
Angle Addition Postulate
Substitution from 19 and 20
Distributive Property of Multiplication
Identity property of division
Substitution from 10 and 15
Consecutive angles of a parallelogram are
supplementary
Identity property of division
Distributive property of multiplication
Substitution from 8 and 15
Substitution from 16 and 24
Substitution from 27, 28, and 29
Triangle Sum
Identity property of subtraction
Substitution from 30 and 32
A right triangle has one angle whose
measure is 90o
Bisectors of Consecutive Angles – Final Part B - Problem 7 – Doug Ruby
We have now proven the following when the consecutive angle bisectors of a parallelogram
intersect on a point on the remaining side of the parallelogram:
1. The ratio of the adjacent sides of the parallelogram must be 2:1.
2. The two smaller triangles ( ADE and BCE) are both isosceles triangles whose
congruent sides are equal to the smaller of the sides of the parallelogram.
3. ADE is an isosceles triangle whose base angles are each ½ of CDA.
4. BCE is an isosceles triangle whose base angles are each ½ of BCD.
5. The larger center triangle ( DEC) is always a right triangle (regardless of whether point
E is on segment AB or not).
I believe that these 5 points summarize key observations (with proof) about the nature of the
parallelogram and the three triangles formed when the angle bisectors of consecutive angles of
the parallelogram intersect on the remaining side of the parallelogram.
Note: Please see the next diagram for an Addendum regarding the areas of the three triangles.
Notice that the area of ADE and BCE are equal to ½ of the area of DEC. A re-examination
of the diagram from our prior proofs will show this more clearly. For this proof, we will use an
analytic paragraph style rather than a straight two-column approach.
Page 7
Bisectors of Consecutive Angles – Final Part B - Problem 7 – Doug Ruby
Addendum: Semi-analytic proof of the areas of the triangles in the Bisectors of Consecutive
Angles Problem
A
F
E
h
h
D
G
B
h
H
C
Notice in the construction above, that we have parallelogram ABCD with base CD and height h.
Thus, the area of this parallelogram is:
1. AABCD = CD × h
Further, since E is on AB by construction and by proof, the height of DEC is h and the area of
DEC is:
2. ADEC = ½ × CD × h
From Proof #1, we observed that point E bisects AB. Since AB = CD (property of
parallelograms):
3. AE = EB = ½ AB = ½ CD
Since AB || CD, we know that the height h of triangles ADE and BCE is the same h as DEC.
The area of ADE and BCE are:
4. AADE = ½ × AE × h
5. ABCE = ½ × EB × h
Substituting from 3, we find that the areas of ADE and BCE are equal and are:
6. AADE = ½ × ½ CD × h = ¼ × CD × h
7. ABCE = ½ × ½ CD × h = ¼ × CD × h
Page 8
Bisectors of Consecutive Angles – Final Part B - Problem 7 – Doug Ruby
Finally, by comparison, we find that:
8.
ADEC
AADE
ADEC
ABCE
1
CD h
2
1
CD h
4
1
2
1
4
1 4
2 1
2
Thus, using the information derived from our two Euclidean proofs, we constructed a simple
analytic “style” paragraph proof that shows that the area of the two smaller triangles ( ADE and
BCE) are equal to ½ the area of the larger triangle ( DEC) in our construction.
Page 9
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