In the figure, $s e g P S ⊥ s i d e Q R$ if $P Q = a, P R = b, Q S = c$ and $R S = d$ , then prove that
$(a + c) (a - c) = (b + d) (b - d)$ .

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Solution

In $Δ P Q S$
$P Q^{2} = P S^{2} + Q S^{2} P S^{2} = a^{2} - c^{2} = (a + c) (a - c) \dots (1)$
In $Δ P R S$
$P R^{2} = P S^{2} + R S^{2} P S^{2} = b^{2} - d^{2} = (b + d) (b - d) \dots (2)$
From $(1) (2)$
$⟹ (a + c) (a - c) = (b + d) (b - d)$

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