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10. 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.
Answer :

Given: Two circles are drawn with sides AC and AB of \(\triangle{ABC}\) as diameters. Both circles intersect each other at D.

To prove: D lies on BC.

Construction: Join AD.
Proof: Since, AC and AB are the diameters of the two circles,
\(\angle{ADB}\) = \(90^\circ\) ...(i)(Angles in a semi-circle)
And \(\angle{ADC}\) = \(90^\circ\) ...(ii)(Angles in a semi-circle)
On adding Equation (i) and (ii), we get,
\(\angle{ADB}\) + \(\angle{ADC}\) = \(90^\circ\) + \(90^\circ\) = \(180^\circ\)
Hence, we can say that, BCD is a straight line.
So, D lies on BC.