# CBSE Class 10 Answered

**In the figure, OP is equal to the diameter of the circle. Prove that ABP is an equilateral triangle.**

Asked by Topperlearning User | 27 Jul, 2017, 01:09: PM

Expert Answer

Say,Radius (OA) = r

_{} OP = 2r………given,OP= diameter of the circle

_{}OAP = 90^{o} (Tangent is _{} to the radius
through the point of contact)

In
right _{}OAP

sin
(_{}OPA) = _{}

_{} _{}OPA = 30^{o}

Similarly
_{}OPB = 30^{o}

_{} _{}APB = 30^{o} + 30^{o} = 60^{o}

Since PA = PB (Lengths of tangents from an external point are equal)

_{} _{}PAB = _{}PBA

In
_{}APB,

_{}APB
+ _{}PAB + _{}PBA = 180^{o} (angle sum property)

_{} 60^{o} + 2_{}PAB = 180^{o}

_{} _{}PAB = 60^{o}

_{} _{}PBA = 60^{o}

Since
all angles are 60^{o}, _{}ABP is equilateral triangle.

Answered by | 27 Jul, 2017, 03:09: PM

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