Question

let BD ∩ CE ( call it A ) such that B*A*D and C*A*E ...
if M is the midpoint of DC ,
< ACM congruent <ADM , <BMC congruent <EMD , then AE = AB

Answer 1

Let BD $\cap$ CE ( call it A ) such that B*A*D and C*A*E ...If M is the midpoint of DC, < ACM congruent <adm ,=""> BMC congruent <emd ,=""> then AE = AB?

Solution:

1) Because $\angle ACM = \angle ADM$ then $CA = DA = b$ by Base Angle Converse.

(http://www.regentsprep.org/Regents/math/geometry/GPB/theorems.htm)

2) Because M is the midpoint of DC then $CM = DM$ .

3) $\angle CME = \angle DMB$ because $\angle CME = (\beta + \gamma)$ and $\angle DMB = (\beta + \gamma)$ .

4) By Angle-Side-Angle Congruence we have

$$\Delta C M E = \Delta D M B.$$

Thus

$$C E = D B,$$ $$C A + A E = D A + A B,$$ $$b + A E = b + A B,$$ $$A E = A B$$

that we'll need to prove.</emd></adm>