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 1 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,

`PQ^2 = PD^2 + QD^2`   .....[By Pythagoras theorem]

⇒ `a^2 = PD^2 + c^2`

⇒ `PD^2 = a^2 - c^2`  .....(i)

In right-angled ∆PDR, 

`PR^2 = PD^2 + DR^2`   .......[By Pythagoras theorem]

⇒ `b^2 = PD^2 + d^2`

⇒ `PD^2 = b^2 - d^2`  .....(ii)

From equations (i) and (iii),

`a^2 - c^2 = b^2 - d^2`

⇒ `a^2 - b^2 = c^2 - d^2`

⇒ `(a - b)(a + b) = (c - d)(c + d)`

Hence proved