SOLUTION: Hello there, I am hoping that you can help me with this geometry proof. I have been looking everywhere for an example, but have had no luck. I even checked geometry books out at ou

Question 720830: Hello there, I am hoping that you can help me with this geometry proof. I have been looking everywhere for an example, but have had no luck. I even checked geometry books out at our local library.
I believe that the problem is based on the Pythagorean Theorem Proof Using Similarity.
Given: ∆ ABC, AD bisects ∠BAC, and AE ≅ ED
Prove: AE/AC = BD/BC
The picture provided shows a triangle labeled ABC. AD bisects angle BAC. From point D there is another line that extends to the side of the triangle labeled AC, this point is labeled E. I am sorry that I am unable to attach a picture. Any help would be appreciated!

Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
You have to prove angle DEC is congruent to angle BAC
( exterior angle theorem)
angle DEC = angle DAE +EDA
But DAE =ADE=BAD ( given)
In triangles ABC & CED angle C is common angle.
DEC = BAC
Triangles are similar.
Take the ratio of sides ( basic proportionality theorem)
EC/AC = DC/BC
(AC-AE)/AC = (BC-BD)/BC
AC/AC - AE/AC = BC/BC- BD/BC
1-AE/AC = 1- BD/BC
therefore AE/AC = BD/BC

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