From a point P, two tangents PA and PB are drawn to a circle with centre O. If OP = diameter of the circle, show that $Δ A P B$ is equilateral.

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Solution

$A P^{2} + O A^{2} = O P^{2} \Rightarrow A P^{2} + r^{2} = (2 r)^{2} \Rightarrow A P^{2} + r^{2} = 4 r^{2} \Rightarrow A P^{2} = 3 r^{2} \Rightarrow A P = \sqrt{3 r} = P B$
Now, $a r (Δ O A P) = \frac{1}{2} \times O A \times A P o r \frac{1}{2} \times A M \times O P \Rightarrow / \frac{1}{2} \times / r \times \sqrt{3 r} = / \frac{1}{2} \times A M \times 2 r$
$\Rightarrow A M = \frac{\sqrt{3}}{2} r ∴ A B = 2 A M = \sqrt{3} r ∴ A P = P B = A B \Rightarrow Δ A P B$ is equilateral $Δ$

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From a point $P$ , two tangents $P A$ and $P B$ are drawn to a circle $C (O, r)$ . If $O P = 2 r$ , show that $△ A P B$ is equilateral.

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