In right-angled triangle PQR, ∠P = 90°, and PS is perpendicular to the hypotenuse QR. Prove that: $$ \frac{1}{PS^2} - \frac{1}{PQ^2} = \frac{1}{PR^2} $$

Q.In right-angled triangle PQR, ∠P = 90°, and PS is perpendicular to the hypotenuse QR. Prove that: $$ \frac{1}{PS^2} - \frac{1}{PQ^2} = \frac{1}{PR^2} $$

Given: PQR is a right-angled triangle with ∠P = 90°, and PS is perpendicular to the hypotenuse QR. To Prove: $$ \frac{1}{PS^2} - \frac{1}{PQ^2} = \frac{1}{PR^2} $$ **Proof:** Since triangles $ \triangle PSR $ and $ \triangle PQS $ are similar, ∴ $ \frac{PS}{RS} = \frac{QS}{PS} $ ⇒ $ PS^2 = RS \cdot QS $ ——— (i) Again, triangles $ \triangle PQR $ and $ \triangle PQS $ are similar, ∴ $ \frac{PQ}{QS} = \frac{QR}{PQ} $ ⇒ $ PQ^2 = QS \cdot QR $ ——— (ii) And triangles $ \triangle PQR $ and $ \triangle PRS $ are similar, ∴ $ \frac{PR}{QR} = \frac{RS}{PR} $ ⇒ $ PR^2 = RS \cdot QR $ ——— (iii) Now, LHS = $ \frac{1}{PS^2} - \frac{1}{PQ^2} $ = $ \frac{1}{RS \cdot QS} - \frac{1}{QS \cdot QR} $ [Using (i) and (ii)] = $ \frac{QR - RS}{RS \cdot QS \cdot QR} $ = $ \frac{QS}{RS \cdot QS \cdot QR} $ = $ \frac{1}{RS \cdot QR} $ = $ \frac{1}{PR^2} $ [Using (iii)] = RHS Proved.