Question

Circles are described on the sides of a triangle as diameters. Prove that the circles on any two sides intersect each other on the third side (or third side produced).

Answer 1

$$\begin{gathered} \angle ADB=90°\left(\mathrm{Angle}\mathrm{in}\mathrm{a}\mathrm{semicircle}\right) \\ \angle ADC=90°\left(\mathrm{Angle}\mathrm{in}\mathrm{a}\mathrm{semicircle}\right) \\ \mathrm{So},\angle ADB+\angle ADC=90°+90°=180° \\ \mathrm{Therefore},BDC\mathrm{is}\mathrm{a}\mathrm{line}. \\ \mathrm{Hence},\mathrm{the}\mathrm{point}\mathrm{of}\mathrm{intersection}\mathrm{of}\mathrm{two}\mathrm{circles}\mathrm{lie}\mathrm{on}\mathrm{the}\mathrm{third}\mathrm{side}. \end{gathered}$$