Question

$P,Q,R$ are, respectively, the mid points of sides $AB,BC$ and $CA$ and of a triangle $ABC$. $PR$ and $AQ$ meet at $X$. $BR$ and $PQ$ meet at $Y$. Prove that $XY=1/4AB$

Answer 1

By mid point theorem

In $△ABC,QR=\frac{1}{2}BC$

$XY\parallel QR$

$XY=\frac{1}{2}QR$

$QR=2xy$

$QR=\frac{1}{2}BA$

$2xy=\frac{1}{2}BA$

$XY=\frac{1}{4}BA$