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

Given:

In \( \Delta \mathrm{PQR}, \mathrm{PD} \perp \mathrm{QR} \) such that \( \mathrm{D} \) lies on \( \mathrm{QR} \).

\( \mathrm{PQ}=a, \mathrm{PR}=b, \mathrm{QD}=c \) and \( \mathrm{DR}=d \)

To do:

We have to prove that \( (a+b)(a-b)=(c+d)(c-d) \).

Solution:

In right angle triangle $\mathrm{PDO}$,

$P Q^{2}=P D^{2}+Q D^{2}$

$a^{2} =P D^{2}+c^{2}$

$P D^{2} =a^{2}-c^{2}$........(i)

In right angle triangle $PDR$,

$P R^{2} =P D^{2}+D R^{2}$

$b^{2}=P D^{2}+d^{2}$

$P D^{2}=b^{2}-d^{2}$..........(ii)

From (i) and (ii), we get,

$a^{2}-c^{2}=b^{2}-d^{2}$

$a^{2}-b^{2}=c^{2}-d^{2}$

$(a-b)(a+b)=(c-d)(c+d)$

Hence proved.