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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.