Question

In $\Delta PQR$, $PD\perp QR$ such that D lies on QR, if PQ = a, PR = b, QD = c and DR = d, then prove that (a+b) (a-b) = (c+d) (c-d).

Answer 1

Given in $\Delta PQR,PD\perp QR,PQ=q,PR=b,QD=c$ and $DR=d$
To prove $(a+b)(a-b)=(c+d)(c-d)$
Proof In right angled $\Delta PDQ$,
$P{Q}^2=P{D}^2+Q{D}^2$ [by Pythagoras theorem]
$\Rightarrow a^2=P{D}^2+c^2$
$\Rightarrow P{D}^2=a^2-c^2$ …..(i)



In right angled $\Delta PDR$, $P{R}^2=P{D}^2+D{R}^2$ [by Pythagoras theorem]
$\Rightarrow b^2=P{D}^2+d^2$
$\Rightarrow P{D}^2=b^2-d^2$ ……(ii)
From Eq.(i) and (ii),
$a^2-c^2=b^2-d^2$
$\Rightarrow a^2-b^2=c^2-d^2$
$\Rightarrow (a-b)(a+b)=(c-d)(c+d)$