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

Given:

From a point \( P \), two tangents \( P A \) and \( P B \) are drawn to a circle with centre \( O \).

\( O P= \) diameter of the circle.

To do:

We have to show that \( \Delta A P B \) is equilateral.

Solution:

Join $AB, OP, AQ, OA$.

Let $r$ be the radius of the circle.

This implies,

$OP = 2r$

$OQ + QP = 2r$

$OQ = QP = r$

In right angled triangle $OAP$,

$OP$ is the hypotenuse and $Q$ is its mid point.

$OA = AQ = OQ$    (Mid-point of hypotenuse of a right angled triangle is equidistant from its vertices)

Therefore,

$\triangle OAQ$ is an equilateral triangle and $\angle AOQ = 60^o$.

In right angled triangle $OAP$,

$\angle APO = 90^o - 60^o = 30^o$

$\angle APB = 2 \angle APO = 2 \times 30^o = 60^o$

$PA = PB$  (Tangents from a point to the circle are equal)

$\angle PAB = \angle PBA = 60^o$

This implies,

$\triangle APB$ is an equilateral triangle.

Hence proved.

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Updated on: 10-Oct-2022

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