Question

In the follwing figure PQRS is a rhombus formed by joining the mid-points of a quadrilateral YMXN show that $3P{Q}^2=S{N}^2+N{R}^2+Q{X}^2+X{R}^2+P{Y}^2+Y{S}^2$

Answer 1

Given that, PQRS is a rhombus formed by joining mid-points of the sides of the quadrilateral YMXR
$\therefore$ YMXR is a rectangle
$\Rightarrow \angle Y=\angle M=\angle X=\angle R={90}^{\circ }$
In triangle $SNR,S{N}^2+N{R}^2=S{R}^2$
In triangle $QXR,Q{X}^2+X{R}^2=Q{R}^2$
In triangle $PYS,P{Y}^2+Y{S}^2=P{S}^2$
Since PQRS is a rhombus, $PQ=RQ=RS=SP$
$S{R}^2+Q{R}^2+P{S}^2=3P{Q}^2$
$\Rightarrow 3P{Q}^2=S{N}^2+N{R}^2+Q{X}^2+X{R}^2+P{Y}^2+Y{S}^2$