Question

In the figure, PS & PT trisect $\Delta$ such that QS=ST=TR. $\angle Q$ is ${90}^{\circ }$. Prove that - $8P{T}^2=3P{R}^2+5P{S}^2$.

Answer 1

Let $QR=3a$

$QS=ST=TR=a$

In $\Delta PQR$

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

$P{R}^2=P{Q}^2+9a^2$..................... (1)

In $\Delta PQT$

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

In $\Delta PQS$

$P{S}^2=P{Q}^2+a^2$......................(2)

$3P{R}^2+5P{S}^2$

$3P{Q}^2+27a^2+5P{Q}^2+5a^2$

$8P{Q}^2+32a^2$

$=8P{T}^2$

$8P{T}^2=3P{R}^2+5P{S}^2$