Question

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).

Answer 1

Given: In ∆PQR,

PD ⊥ QR,

PQ = a,

PR = b,

QD = c 

And DR = d

To prove: (a + b)(a – b) = (c + d)(c – d)

Proof: In right angled ΔPDQ,

PQ² = PD² + QD²   ...[By pythagoras theorem]

⇒ a² = PD² + c²

⇒ PD² = a² – c²  ...(i)

In right angled ∆PDR, 

PR² = PD² + DR²   ...[By pythagoras theorem]

⇒ b² = PD² + d²

⇒ PD² = b² – d²  ...(ii)

From equations (i) and (ii),

a² – c² = b² – d²

⇒ a² – b² = c² – d²

⇒ (a – b)(a + b) = (c – d)(c + d)

Hence proved.