Question

ABC and ADC are two right triangles with common hypotenuse AC. Prove that $\angle \mathrm{CAD}=\angle \mathrm{CBD}$.

Answer 1

Proving $\angle \mathrm{CAD}=\angle \mathrm{CBD}$:

From figure, it is clear that,

$\angle \mathrm{ABC}=\angle \mathrm{ADC}={90}^{\circ }$, which means

These angles are in the semicircle.

Therefore both triangles $∆\mathrm{ABC}$ and $∆\mathrm{ADC}$ are lying in the semicircle.

AC is the diameter of circle with center O.

Therefore, Points A, B, C and D are concyclic means all the points lie on the circle.

Thus, CD is a chord.

Therefore, $\angle \mathrm{CAD}=\angle \mathrm{CBD}$ (Angles in the same segment of the circle)

Hence Proved.