Question

Plzz refer the image for given and to prove

Answer 1

Since $\sf{OC}$ is a radius of the circle and $\sf{CP}$ is a tangent to the circle,

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$\sf{\angle OCP=90^{\circ}}$

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And we see that,

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$\sf{OP=OA+AP}$

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Since $\sf{OA=AP,}$

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$\sf{OP=2OA}$

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But since $\sf{OA}$ and $\sf{OC}$ are radii of the same circle,

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$\sf{OP=2OC}\\\\\\\sf{\dfrac {OC}{OP}=\dfrac {1}{2}}$

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Then, in $\sf{\triangle OCP,}$

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$\sf{\cos\angle POC=\dfrac {OC}{OP}}\\\\\\\sf{\cos\angle POC=\dfrac {1}{2}}$

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This implies,

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$\sf{\angle POC=60^{\circ}}$

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since $\sf{\angle POC}$ is acute.

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The quadrilateral $\sf{AOCE}$ is cyclic because $\sf{AE}$ and $\sf{CE}$ are tangents to the circle. Then,

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$\sf{\angle CED=\angle AOC=60^{\circ}}$

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And in $\sf{\triangle BOC,\ OB=OC.}$

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$\sf{\therefore\ \angle OBC=\angle OCB=\dfrac {\angle AOC}{2}=30^{\circ}}$

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because we know that sum of two interior angles of a triangle is equal to the third exterior angle.

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Then,

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$\sf{\angle ECD=180^{\circ}-(\angle OCB+\angle OCP)}\\\\\\\sf{\angle ECD=60^{\circ}}$

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That is,

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$\sf{\angle ECD=\angle CED=60^{\circ}}$

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This implies $\sf{\triangle CDE}$ is nothing but an equilateral triangle.

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Hence Proved!