Question

In an isosceles triangle PQR, PR = QR and PS is perpendicular to QR. Prove that PS2 – QS2 = 2QS.RS.
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Answer 1

$\textbf{Given:}$

$\text{In \triangle PQR, PR=QR and PS{\perp}QR }$

$\textbf{To prove:}$

$PS^2-QS^2=2\,QS\,RS$

$\textbf{Solution:}$

$\text{In \triangle PSR,}$

$\text{By pythagoras theorem,}\bf\;PR^2=PS^2+RS^2$-----------(1)

$\text{Consider,}$

$PS^2-QS^2$

$\text{Using (1)}$

$=(PR^2-RS^2)-(QR-RS)^2$

$=(PR^2-RS^2)-(QR^2+RS^2-2\,QR\,RS)$

$=PR^2-RS^2-QR^2-RS^2+2\,QR\,RS$

$\text{But,}\;PR=QR$

$=PR^2-RS^2-PR^2-RS^2+2\,QR\,RS$

$=2\,QR\,RS-2\,RS^2$

$=2\,RS(QR-RS)$

$=2\,RS\,QS$

$\implies\boxed{\bf\,PS^2-QS^2=2\,QS\,RS}$

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