Question

32. In the given figure, OP is equal to the diameter of a circle with centre O and PA and PB are tangents.
Prove that ABP is an equilateral triangle.
A
PO
o
Q
B​

Answer 1

PA and PB are the tangents to the circle.

∴ OA⊥PA

⇒ ∠OAP=90°

In ΔOPA,

sin∠OPA=

OP

OA

=

2r

r

[Given OP is the diameter of the circle]

⇒ sin∠OPA=

2

1

=sin30⁰

⇒ ∠OPA=30°

Similarly, it can be proved that ∠OPB=30°.

Now, ∠APB=∠OPA+∠OPB=30°+30°=60°

In ΔPAB,

PA=PB [lengths of tangents drawn from an external point to a circle are equal]

⇒∠PAB=∠PBA....(1) [Equal sides have equal angles opposite to them]

∠PAB+∠PBA+∠APB=180° [Angle sum property]

⇒∠PAB+∠PAB=180°–60°=120° [Using (1)]

⇒2∠PAB=120°

⇒∠PAB=60° .(2)

From (1) and (2)

∠PAB=∠PBA=∠APB=60°

∴ ΔPAB is an equilateral triangle.

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