Question

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 1

Prove the given statement

Consider $AB$ and $AC$ are the diameters of two circle

Then we have to prove that the point of intersection $D$ lie on $BC$

Let circles with $AB$ and $AC$ as diameter intersect at $D$

Then $\angle ADB=\angle ADC=90°[\mathrm{Angle}\mathrm{in}\mathrm{a}\mathrm{semicircle}]$

Now,

$\Rightarrow \angle ADB+\angle ADC=180°[\mathrm{Linear}\mathrm{pair}]$

This implies $B,D,C$ lie on the same line.

Hence both the circles meet the third side at $D$

Hence proved that the point of intersection of these circles lie on the third side.