Question

$P,Q$ and $R$ are, respectively, the mid points of sides $BC,AB$ of a triangle $ABC,PR$ and $BQ$ meet at $X.CR$ and $PQ$ meet at $Y$. Prove that $XY=\frac{1}{4}BC$

Answer 1

In a triangle line joining mid - point of two sides are parallel and half of third side

$\begin{array}{c}PQ\parallel AB\\ PQ=\frac{1}{2}AB\to \left(i\right)\\ QR\parallel BC\\ QR=\frac{1}{2}BC\to \left(ii\right)\\ PR\parallel AC\\ PQ=\frac{1}{2}AC\to \left(iii\right)\\ fromequation\left(i\right)\left(ii\right)\end{array}$

two pairs are parallel so, BPQR is a parallelogram

$BP=QR=\frac{BC}{2}\to \left(iv\right)$

In parallelogram diagonals bisects each other so

X is mid -point to PR. Similarly,

Y is mid- point on QP

$\begin{array}{c}In\Delta RPQ\\ XY=\frac{1}{2}QR\\ XY=\frac{1}{2}\left(\frac{1}{2}BC\right)\\ \frac{1}{4}BC\end{array}$