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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:
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.

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.