Question

In fig-10, $△ABC$ and $△ADC$ are two right triangles with common hypotenuse AC. Show that, $\angle CAD=\angle CBD$ .

Answer 1

The triangles formed on diameter of circle and the vertex on the circle have same area and same height

$⟹AC||BD$

and $AD$ is transversal

The Interior Alternate angles are equal

$⟹\angle CAD=\angle ADB$