Question

In the given figure, XY and XY" are two parallel tangents to a circle with centre O and another tangent
AB with point of contact C, is intersecting XY at A and X'Y'at B. Prove that angle AOB = 90°​

Answer 1

$\huge{\boxed{\mathtt{\red{ANSWER}}}}$

(Mark as brainlist)

Answer 2

Solution: In ∆POA and ∆COA we get,

→ AP = AC [Tangent]

→ OA = OA [Common]

→ OC = OP [Radius]

Hence ∆POA ≈ ∆COA by SSS congruency.

→ ∠POA = ∠COA [c.p.c.t]

Similarly, ∆BOC ≈ ∆BOQ by SSS congruency.

→ ∠BOC = ∠BOQ [c.p.c.t]

Now ∠BOQ + ∠BOC + ∠AOC + ∠POA = 180°

→ ∠BOC + ∠BOC + ∠AOC + ∠AOC = 180°

→ 2(∠BOC + ∠AOC) = 180°

→ ∠AOB = 90°

Hence Proved