Two tangent segment PA and PB are drawn to a circle center O such that angle APB = 120. Prove that OP = 2 AP.

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Solution

Triangles PAO and PBO can be proved congruent using RHS criterion.
Thus, $∠ A P O = ∠ B P O$ (CPCT)
Given that $∠ A P B = 120^{\circ}$
$A P B = A P O + B P O = 2 A P O = 120$
$A P O = 60^{\circ}$
In triangle APO
$c o s 60 = \frac{1}{2} = \frac{A P}{O P}$
Thus, OP = 2AP
Hence Proved

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Similar questions

Two tangent segments PA and PB are drawn to a circle with centre O such that $∠ A P B = 120^{\circ}$ . Prove that OP = 2AP.

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∠ A P B = 120^{\circ}

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Two tangents PA and PB are drawn to the circle with centre O, such that $∠$ APB $= 120^{\circ}$ . What is the relation between OP and AP?

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