Given : PD ⊥ QR. PQ = a, PR = b, QD and DR = d.
To prove : (a + b)(a – b) = (c + d)(c – d)
Proof : In right ∆PDQ
PQ2 = QD2 + PD2 (By Pythagoras Theorem)
⇒ PD2 = PQ2 – QD2 ….(i)
In right ∆PDR
PR2 = DR2 + PD2
⇒ PD2 = PR2 – DR2 …..(ii)
From equation (i) and (ii)
PQ2 – QD2 = PR2 – DR2
⇒ a2 – c2 = b2 – d2 [putting values of PQ, QD, PR and DR]
⇒ a2 – b2 = c2 – d2
⇒ (a + b)(a – b) = (c + d)(c – d).