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.
Try using the angle bisector theorem and similar triangles to prove equality of distances. From there, you are right about the parallelogram. One thing though is I know this statement is true for medians, but might be false for angle bisectors.
To do this, we use a couple of theorems.
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 $DE || BC$.