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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.