Question

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

Open in App
Solution

## Proving $\angle \mathrm{CAD}=\angle \mathrm{CBD}$:From figure, it is clear that,$\angle \mathrm{ABC}=\angle \mathrm{ADC}={90}^{\circ }$, which meansThese 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.

Suggest Corrections
0
Explore more