Question

❋ 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 angle AOB =90°

❋ Please answer it correctly.
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Answer 1

Refer The Attachment ⬆️

Hence Proved, Angle AOB = 90°

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Answer 2

$\large\bf\underline\red{Answer \: :- }</h3><h3>$

Construction :- Join O and C

In ∆OPA and ∆OCA

OP = OC ( radii of the same circle )

AP = AC ( tangent form the point A )

AO = AO ( common side )

∆OPA ≈ ∆OCA ( SSS congruence criterion )

Therefore,

$\sf{P \longleftrightarrow C }$

$\sf{A \longleftrightarrow A }</p><p>$

$\sf{O \longleftrightarrow O }</p><p>$

$\sf{ \angle POA = \angle COA \: \: \: \: ...(1) }$

By similarity,

$\sf{ \angle QOB \cong \angle OCB }$

$\sf{ \angle QOB = \angle COB \: \: \: \: ..(2)}$

Since,

POQ is the diameter of the circle, it is a straight line.

Therefore,

$\small\sf{ \angle POA + \angle COA + \angle COB + \angle QOB = 180° }$

So, from equation (1) and equation (2)

$\longmapsto\sf{ 2 \angle COA + 2 \angle COB = 180°} \\ \\ \longmapsto\sf{ \angle COA + \angle COB = 90°} \: \: \: \: \: \: \: \\ \\ \longmapsto\sf{ \angle AOB = 90°}$

Hence Proved

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