In △PQR, PD ⊥ 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).

In right angle △PDQ,

By pythagoras theorem we get,

⇒ PQ² = PD² + QD²

⇒ a² = PD² + c²

⇒ PD² = a² - c² ……..(i)

In right angle △PDR,

By pythagoras theorem we get,

⇒ PR² = PD² + DR²

⇒ b² = PD² + d²

PD² = b² - d² ……..(ii)

From (i) and (ii) we get,

⇒ a² - c² = b² - d²

⇒ a² - b² = c² - d²

⇒ (a - b)(a + b) = (c - d)(c + d).

Hence, proved that (a - b)(a + b) = (c - d)(c + d).