If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.

Given:

Circles are drawn taking two sides of a triangle as diameters.

To do:

We have to prove that the point of intersection of these circles lie on the third side.

Solution:

Let us draw a triangle $PQR$ and two circles having diameters as $PQ$ and $PR$ respectively.

We know that,

Angles in a semi-circle are equal.

This implies,

$\angle PSQ = \angle PSR = 90^o$

$\angle PSQ+\angle PSR = 180^o$

Therefore,

$\angle BDC$ is a straight line.

So, we can say that $S$ lies on the line $QR$.

Hence proved.

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Updated on: 10-Oct-2022

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