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.
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$\begingroup$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.
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