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

(N/A) Given: In $\triangle PQR$,$PD \perp QR$,$PQ = a$,$PR = b$,$QD = c$,and $DR = d$.
To prove: $(a+b)(a-b) = (c+d)(c-d)$.
Proof: In right-angled $\triangle PDQ$,by Pythagoras theorem:
$PQ^2 = PD^2 + QD^2$
$a^2 = PD^2 + c^2$
$PD^2 = a^2 - c^2$ ...... $(i)$
In right-angled $\triangle PDR$,by Pythagoras theorem:
$PR^2 = PD^2 + DR^2$
$b^2 = PD^2 + d^2$
$PD^2 = b^2 - d^2$ ...... $(ii)$
From equations $(i)$ and $(ii)$:
$a^2 - c^2 = b^2 - d^2$
$a^2 - b^2 = c^2 - d^2$
$(a - b)(a + b) = (c - d)(c + d)$
Hence,$(a + b)(a - b) = (c + d)(c - d)$ is proved.