Parallelogram proof problem

1. Dec 4, 2009

1/2"

1. The problem statement, all variables and given/known data
ABCD is a paralleelogram.angle DAE=angleEAB and angle CBE=angleEBA .Prove that AB=2BC

2. Relevant equations

none

3. The attempt at a solution
I just got that angle AEB =90 degrees
(But it isn't of any help)

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Last edited: Dec 4, 2009
2. Dec 4, 2009

1/2"

Re: Question

Hey I have worked on a solution but i dont know if it's correct.
If i extend the line AE into EF such that FB=BC
and if angle EAB=y=angle FAD and angle CBA=2x
=> 2x+y=angleFBA+2x =>y=angleFBA
.: As angleFAB= angle AFB
.: AB = FB=>
=>AB=2BC
Is it correct? Please let me know if it is wrong.

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Last edited: Dec 4, 2009
3. Dec 5, 2009

Staff: Mentor

Re: Question

From your second drawing, FB can't possibly be equal to BC. Maybe you didn't say what you meant, but what you said was incorrect.
In your first post you established that angle AEB = 90 deg. You didn't think that was any help, but it is. I worked this problem by labeling all of the interior angles of your original figure.

You are given that DAE = EAB. Let a = the measure of each of these angles. Also, since ABE = EBC, let the measure of each of these angles = b. Since opposite angles of a parallelogram are congruent, ADE = 2b and BCE = 2a.

Now let the measure of DEA = c and the measure of BEC = d. From the fact that the sum of the interior angles of a triangle is 180 deg., you can find c and d, and determine that triangle ADE is isosceles and that triangle BEC is also isosceles. You can also determine that these two triangles are congruent, and once you do that, you can show that AB = 2BC.