# Prove that DE || BC

Let M be the midpoint of side BC in triangle ABC. The angle bisector of BMA intersects AB in D, while the angle bisector of CMA intersects AC in E. How can i prove that DE||BC? I drew out the triangle and all the bisectors and points but I have no idea where to start. I was thinking maybe I can prove CE=MD and then by parallelogram DE||BC? I have no idea how to go about this though. Any help would be appreciated.

So the first is the angle bisector theorem. This says that in the triangle $ABM$, the angle bisector $DM$ divides the line $AB$ in the ratio $BM:MA$, which is to say $BD:DA=BM:MA$. On the other triangle, a similar principle is applied to get $CE:EA=CM:MA=BM:MA$ as $M$ is the midpoint of $BC$.
Therefore $CE:EA=BM:MA=BD:DA$. Adding $1$ to the left and right sides of this equality (treated as a fraction), we get $AC:AE=AB:AD$, or transposing,$AC:AB=AE:AD=k$ (say).
Looking at the triangles $ABC$ and $ADE$, they share a common angle at $A$ ,and the two sides not opposite the angle are proportional. Now, we write:$BC^2=AB^2+AC^2-2AB\cdot AC \cos A = k^2(AD^2+AE^2+2AD \cdot AE \cos A) = k^2DE^2 \implies BC=k.DE$. Thus the triangles $ABC$ and $ADE$ are similar, and thus $\angle ADE=\angle ABC$. But these are corresponding angles in that case, and that gives $DE || BC$.