In $\triangle\text{PQR},\text{PD}\perp\text{QR}$ such that D lies on QR. If PQ = a, PR = b, QD = C and DR = d, prove that (a + b)(a - b)=(a + b)(c - d).
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Given: In $\triangle\text{PQR},\text{PD}\perp\text{QR}$ so $\angle1=\angle2$
PQ = a, PR = b, QD = C and DR = d,
To prove: (a + b)(a - b) = (c + d)(c - d)
prove: In right angle $\triangle\text{PQR},$
$PD^2 = PQ^2 - QD^2$ [by Pythagoras theorem]
$\Rightarrow PD^2 = a^2 - c^2 .......(i)$
Similarly, in right angled $\triangle\text{PQR},$
$PD^2 = PR^2 - DR^2$
$\Rightarrow PD^2 = b^2 - a^2 ......(ii)$
From (i) and (ii), we have
$a^2- c^2 = b^2 - d^2$
$\Rightarrow a^2 - b^2 = c^2 - d^2$
$\Rightarrow (a - b)(a + b) = (c - d)(c + d)$
Hence, proved.
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