Parallelogram Bisectors Geometry Final Part B Problem 7

Transcription

1 Parallelogram Bisectors Geometry Final Part B Problem 7 By: Douglas A. Ruby Date: 11/10/2002 Class: Geometry 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

2 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 2 nd 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

3 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

4 Statement Reason 1. Parallelogram ABCD Given 2. Point E on AB Given 3. DE bisects CDA, CE bisects BCD Given 4. AD BC Opposite sides of a parallelogram are 5. AB CD Opposite sides of a parallelogram are 6. ADE CDE Angle bisector divides an angle into two congruent angles 7. DCE BCE Angle bisector divides an angle into two congruent angles 8. AED CDE Alternate interior angles 9. AED ADE Substitution from 6 and ADE is isosceles A triangle with two congruent base angles is isosceles 11. BEC DCE Alternate interior angles 12. BEC BCE Substitution from 7 and BCE is isosceles A triangle with two congruent base angles is isosceles 14. AD AE Sides opposite base angles of an isosceles triangle are congruent 15. BC BE Sides opposite base angles of an isosceles triangle are congruent 16. AD BC Opposite sides of a parallelogram are congruent 17. AD BE Substitution from 15 and AE BE Transitive property from 14 and Point E bisects AB Segment bisector is a point that divides a segment into two equal halves 20. AB = AE + BE Segment addition postulate 21. AB = AD + AD Substitution from 14, 17, and AB = 2 AD 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

5 Consecutive Bisectors of a Parallelogram Proof #2 of the Bisectors of the Consecutive Angles of a Parallelogram Theorem: Given: 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. 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

6 Statement Reason 1. Parallelogram ABCD Given 2. Point E on AB Given 3. DE bisects CDA, CE bisects BCD Given 4. Point E bisects AB From Proof #1 5. AB = 2 AD From Proof #1 6. ADE and BCE are isosceles From Proof #1 7. ADE CDE Angle bisector divides an angle into two congruent angles 8. CDE AED Alternate interior angles 9. ADE AED Substitution from 7 and m ADE = m AED Congruent angles have equal measure 11. m CDA = m ADE + m AED Angle Addition Postulate 12. m CDA = m ADE + m ADE Substitution from 10 and m CDA = 2 m ADE Distributive Property of Multiplication 14. m CDA/2 = m ADE Identity property of division 15. m ADE = m AED = m CDA/2 Substitution from 10 and DCE BCE Angle bisector divides an angle into two congruent angles 17. BEC DCE Alternate interior angles 18. BCE BEC Substitution from 16 and m BCE = m BEC Congruent angles have equal measure 20. m BCD = m DCE + m BCE Angle Addition Postulate 21. m BCD = m BCE + m BCE Substitution from 19 and m BCD = 2 m BCE Distributive Property of Multiplication 23. m BCD/2 = m BCE Identity property of division 24. m BCE = m BEC = m BCD/2 Substitution from 10 and m CDA + m BCD = 180 o Consecutive angles of a parallelogram are supplementary 26. (m CDA + m BCD)/2 = 180 o /2 = 90 o Identity property of division 27. m CDA/2 + m BCD/2 = 90 o Distributive property of multiplication 28. m CDE = m CDA/2 Substitution from 8 and m DCE = m BCD/2 Substitution from 16 and m CDE + m DCE = 90 o Substitution from 27, 28, and m CDE + m DCE + m DEC = 180 o Triangle Sum 32. m DEC = 180 o (m CDE + m DCE) Identity property of subtraction 33. m DEC= 180 o 90 o = 90 o Substitution from 30 and DEC is a right triangle A right triangle has one angle whose measure is 90 o Final Part B Problem 7 Doug Ruby - Page 6

7 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

8 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 B h h h D G 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. A ABCD = 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. A DEC = 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. A ADE = ½ AE h 5. A BCE = ½ EB h Substituting from 3, we find that the areas of ADE and BCE are equal and are: 6. A ADE = ½ ½ CD h = ¼ CD h 7. A BCE = ½ ½ CD h = ¼ CD h Page 8

9 Bisectors of Consecutive Angles Final Part B - Problem 7 Doug Ruby Finally, by comparison, we find that: 1 1 CD h ADEC ADEC A 1 1 ADE ABCE 2 1 CD h 4 4 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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