Question

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

Answer 1

Answer:

Join OC

In Δ OPA and Δ OCA

OP = OC (radii of same circle)

PA = CA (length of two tangents)

AO = AO (Common)

Therefore, Δ OPA ≅ Δ OCA (By SSS congruency criterion)

:

Hence, ∠ 1 = ∠ 2 (CPCT)

Similarly ∠ 3 = ∠ 4

Now, ∠ PAB + ∠ QBA = 180°

2∠2 + 2∠4 = 180°

∠2 + ∠4 = 90°

∠AOB = 90° (Angle sum property)

Answer 2

Answer:

Join OC

In Δ OPA and Δ OCA

OP = OC (radii of same circle)

PA = CA (length of two tangents)

AO = AO (Common)

Therefore, Δ OPA ≅ Δ OCA (By SSS congruency criterion)

:

Hence, ∠ 1 = ∠ 2 (CPCT)

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Step-by-step explanation: