Question

In the figure E and F are the midpoints of sides AB and AC of $\Delta ABC$ P is any point on BC. Prove that EF bisects AP.

Answer 1

Since E ,F are the mid points of AB and BC. So , By mid point Theorem,
EF || BC , EF = $\frac{1}{2}$ BC
Now In triangle ABP
Since E is the mid point of AB , EQ || BP , So by Converse of mid point theorem
Q is the mid point of AP
AQ = QP