Question

$P$ and $Q$ are the mid points on the sides $CA$ and $CB$ respectively of triangle $ABC$ right angled at $C$. Prove that $4(A{Q}^2+B{P}^2)=5A{B}^2$

Answer 1

In a right-angled triangle $\Delta ACQ$

$A{Q}^2=A{C}^2+C{Q}^2$

$A{Q}^2=A{C}^2+{\left(\frac{BC}{2}\right)}^2$ ...... $[\because CQ=\frac{BC}{2}]$

$4A{Q}^2=4A{C}^2+(BC{)}^2$ ....... $\left(1\right)$

In a right-angled triangle $\Delta PCB$

$P{B}^2=P{C}^2+B{C}^2$

$P{B}^2={\left(\frac{AC}{2}\right)}^2+B{C}^2$ ....... $[\because PC=\frac{AC}{2}]$

$4P{B}^2=(AC{)}^2+4B{C}^2$....... $\left(2\right)$

Adding $\left(1\right)$ and $\left(2\right)$, we get

$4A{Q}^2+4P{B}^2=4A{C}^2+(BC{)}^2+(AC{)}^2+4B{C}^2$

$4A{Q}^2+4P{B}^2=5A{C}^2+5B{C}^2$

$4A{Q}^2+4P{B}^2=5(A{C}^2+B{C}^2)$

But $A{C}^2+B{C}^2=A{B}^2$ $[\because △ABC$ is a right angled triangle at C$]$

$\therefore 4A{Q}^2+4P{B}^2=5A{B}^2$ $\left[henceproved\right]$