Question 802800: I am having a hard time understanding what I am doing wrong. I am not good at proofs at all. I will try to draw what I am doing since I can not put the picture on here.
Given, Triangle ABC, AD bisects
Prove AE/AC = BD/BC
A
E
B D C
I tried to draw the triangle on here, but don't know how.
A is angle at top
E is below A on the right side
B, D, and C are the bottom angles of the triangle and E bisect A and C
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
I think that you are trying to say that E is the midpoint of AC and D is the midpoint of BC. If this is the case, DE has to be parallel to AB, you can look up the theorem for yourself. That means that angle BAC is congruent to angle DEC by one of the lemmas to the theorem for parallel lines and a transversal, and similarly angle ABC has to be congruent to angle EDC. Then, since angle C is congruent to angle C by reflexive equality, triangle ABC is similar to triangle EDC. The required proportions follow from the properties of similarity.
John
Egw to Beta kai to Sigma
My calculator said it, I believe it, that settles it
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