Question

In the figure below, $S$ and $T$ trisect the side $QR$ of a right triangle $PQR$. Prove that $8P{T}^2=3P{R}^2+5P{S}^2$.

Answer 1

Let $QS=a$

So $QT=2a$ & $QR=3a$

In $△lePQR$

$P{R}^2=P{Q}^2+Q{R}^2$

$P{R}^2=P{Q}^2+(3a{)}^2$

$P{R}^2=P{Q}^2+9a^2$

In $△lePQT$

$P{T}^2=P{Q}^2+4a^2$

In $△lePQS$

$P{S}^2=P{Q}^2+a^2$

So :- $3P{R}^2+5P{S}^2=3P{Q}^2+27a^2+5P{Q}^2+5a^2$

$=8[P{Q}^2+4a^2]$

$=8P{T}^2$