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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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