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

Given,

Two circles are drawn on the sides $AB$ and $AC$ of the triangle

$ABC$ as diameters. The circles intersected at $D$.

Construction: $AD$ is joined.

To prove: $D$ lies on $BC$. We have to prove that $BDC$ is a straight line.

Proof:

$\angle ADB=\angle ADC=90°$ ...Angle in the semi circle

Now,

$\angle ADB+\angle ADC=180°$

$\Rightarrow \angle BDC$ is straight line.

Thus, $D$ lies on the $BC$.