The perpendicular PS on the base QR of a triangle PQR intersects QRat S so that QS=3RS. Prove that 2PQ^2 = 2PR^2 + QR^2

Answer :

We form our diagram from given information , As  :

Here

QS  =  3 RS  ( Given  )

and

QR  =  QS  +  RS  =  3 RS  +  RS  =  4 RS  , So

RS  =  $\frac{QR}{4}$                                             ---- ( A )

Now we apply Pythagoras theorem in triangle PQS and get 

PQ² =  PS² + QS² 

PS² =  PQ² - QS²                                  ---- ( 1 )

And apply Pythagoras theorem in triangle PRS and get 

PR² =  PS² + RS² 

PS² =  PR² - RS²  , Substitute value from equation 1 we get 

$\Rightarrow$PQ² - QS²  =  PR² - RS² 

$\Rightarrow$PQ² =  PR² - RS² + QS²  

$\Rightarrow$PQ² =  PR² - RS² + ( QR - RS )²  

$\Rightarrow$PQ² =  PR² - RS² +  QR² + RS² - 2 QR $\times$RS

$\Rightarrow$PQ² =  PR² +  QR² - 2 QR $\times$RS

Now substitute value from equation A and get

$\Rightarrow$PQ² =  PR² +  QR² - 2 QR $\times$$\frac{QR}{4}$

$\Rightarrow$PQ² =  PR² +  QR² -  $\frac{Q{R}^2}{2}$

$$\begin{gathered} \Rightarrow {\mathrm{PQ}}^2 = \frac{2 {\mathrm{PR}}^2 + 2 {\mathrm{QR}}^2 - {\mathrm{QR}}^2}{2} \\ \Rightarrow {\mathrm{PQ}}^2 = \frac{2 {\mathrm{PR}}^2 + {\mathrm{QR}}^2 }{2} \\ \mathbf{\Rightarrow }\mathbf{2}\mathbf{ }{\mathbf{PQ}}^{\mathbf{2}}\mathbf{ }\mathbf{=}\mathbf{2}\mathbf{ }{\mathbf{PR}}^{\mathbf{2}}\mathbf{ }\mathbf{+}\mathbf{ }{\mathbf{QR}}^{\mathbf{2}}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{(}\mathbf{ }\mathbf{Hence}\mathbf{ }\mathbf{proved}\mathbf{ }\mathbf{)} \end{gathered}$$