Question

If $P$ and $Q$ are point on sides $CA$ and $CB$ respectively of $\Delta ABC$, right angled at $C$, prove that $(A{Q}^2+B{P}^2)=(A{B}^2+P{Q}^2).$

Answer 1

Consider $(A{Q}^2+B{P}^2)$

From $\Delta ACQ$

$A{Q}^2=A{C}^2+C{Q}^2\cdots \left(1\right)$

From $\Delta BPC$

$B{P}^2=C{P}^2+B{C}^2\cdots \left(2\right)$

From $\Delta QPC$

$P{Q}^2=C{P}^2+C{Q}^2\cdots \left(3\right)$

From $\Delta ABC$

$A{B}^2=C{A}^2+C{B}^2\cdots \left(4\right)$

From $\left(1\right),\left(2\right)⟹A{Q}^2+B{P}^2=A{C}^2+C{Q}^2+C{P}^2+B{C}^2$

From $\left(3\right),\left(4\right)⟹A{B}^2+P{Q}^2=A{C}^2+C{Q}^2+C{P}^2+B{C}^2$

$\therefore$ $A{Q}^2+B{P}^2=A{B}^2+P{Q}^2$

Hence proved.