Question

From the external point P, two tangents PA and PB are drawn on the circle with center O.

Prove that ΔABP is an equilateral triangle.
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Answer 1

Step-by-step explanation:

AP is the tangent to the circle.

∴ OA ⊥ AP (Radius is perpendicular to the tangent at the point of contact)

⇒ ∠OAP = 90º

In Δ OAP,

sin ∠OPA = OA/OP = r/2r [Diameter of the circle]

∴ sin ∠OPA = 1/2 = sin 30º

⇒ ∠OPA = 30º

Similarly, it can he prayed that ∠OPB = 30

How, LAPB = LOPP + LOPB = 30° + 30° = 60°

In APPB,

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

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

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

∠PAB + ∠PBA + ∠APB = 180° - 60° [Using (1)]

⇒ 2∠PAB = 120°

⇒ ∠PAB = 60°

From (1) and (2)

∠PAB = ∠PBA = ∠APB = 60°

APPB is an equilateral triangle

Answer 2

Step-by-step explanation:

From the external point P, two tangents PA and PB are drawn on the circle with center O.

Prove that ΔABP is an equilateral triangle