Question

In a $\Delta ABC,$ E and F are the mid-points of AC and AB respectively. The altitude AP to BC intersects FE at Q. Prove that AQ = QP.

Answer 1

Given : In $\Delta ABC$, E and F are the mid-points of AC and AB respectively. EF are joined. $AP\perp BC$ is drawn which intersects EF at Q and meets BC at P.

E and F are the mid points of AC and AB

$\therefore EF||BCandEF=\frac{1}{2}BC$

$\therefore \angle F=\angle B$

In $\Delta ABP$,

F is mid point of AB and Q is the mid point of FE or FQ || Bc

$\therefore$ Q is mid point of AP,

$\therefore$ AQ = QP