Prove basic Parallel line

Question

Given the triangle ABC and [AD] is a median to the side [BC] and P$\in$ [AD]. CP intersects [AB] at E and BP intersects [AC] at F. How to prove that for all possible [EF] are parallel to [BC]? My idea is to extend ABC and add some point to transform it into parallelogram and so it looks a bit easy to prove it. But I think that I miss some part. What should I add to it? How do you prove that?

Answer 1

By Ceva's theorem, $$1 = \frac{AE}{EB} \cdot \frac{BD}{DC} \cdot \frac{CF}{FA} = \frac{AE}{EB} \cdot 1 \cdot \frac{FC}{AF}.$$

Therefore, $$\frac{AE}{EB} = \frac{AF}{FC}.$$

As a consequence, the two triangles $AEF$ and $ABC$ are similar, with $\angle AEF = \angle ABC$ and $\angle AFE = \angle ACB$. From this, the fact that $BC \parallel EF$ follows.