Question

In fig. 6, l and m are two parallel tangents to a circle with centre O, touching the circle at A and B respectively. Another tangent at C intersects the line l at D and m at E. Prove that ∠DOE = 90° ?

Answer 1

Given: l and m at are two parallel tangents to the circle with centre O touching the circle at A and B respectively. DE is a tangent at the point C, which intersects l at D and m at E.

To prove: ∠ DOE = 90°

Construction: Join OC.

Proof:

In ΔODA and ΔODC,

OA = OC (Radii of the same circle)

AD = DC (Length of tangents drawn from an external point to a circle are equal)

DO = OD (Common side)

ΔODA ≅ ΔODC (SSS congruence criterion)

∴ ∠DOA = ∠COD … (1) (C.P.C.T)

Similarly, ΔOEB ≅ ΔOEC

∠EOB = ∠COE … (2)

AOB is a diameter of the circle. Hence, it is a straight line.

∴ ∠DOA + ∠COD + ∠COE + ∠EOB = 180º

From (1) and (2), we have

2∠COD + 2 ∠COE = 180º

⇒ ∠COD + ∠COE = 90º

⇒ ∠DOE = 90°

Hence, proved.