Question

M and N are the mid-points of the sides QR and PQ respectively of a $\Delta$PQR, right-angled at Q. Prove that $P{M}^2+R{N}^2=5M{N}^2$

Answer 1

Construct a line to joining point N to M

$\begin{array}{c}In\Delta PQM\\ P{Q}^2+Q{M}^2=P{M}^2....\left(1\right)\\ In\Delta NQR\\ N{Q}^2+Q{R}^2=N{R}^2...\left(2\right)\\ andIn\Delta NQM\\ N{Q}^2+Q{M}^2=N{M}^2....\left(3\right)\\ Addingeq\left(1\right)and\left(2\right)\\ P{M}^2+N{R}^2=\left(P{Q}^2+Q{R}^2\right)+\left(Q{M}^2+Q{N}^2\right)\\ P{M}^2+N{R}^2={\left(2QN\right)}^2+{\left(2QM\right)}^2+N{M}^2\\ \left[\because MandNismidpointPQandQRrespectively\right]\\ P{M}^2+N{R}^2=4\left(Q{N}^2+4Q{M}^2\right)+N{M}^2\\ fromequation\left(3\right)\\ P{M}^2+N{R}^2=4N{M}^2+N{M}^2=5N{M}^{2}oved\end{array}$