Question

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

Answer 1

Given: $△ABC$ and $△ADC$ are right angled triangles having common hypotenuse $AC$.

To prove : $\angle CAD=\angle CBD$

Proof : In $△ABC,$ we have

$\angle ABC={90}^{\circ }$ [$△ABC$ is a right angled triangle at $B$]

$\angle ABC+\angle BAC+\angle BCA={180}^{\circ }$ [Angle sum property of a triangle]

$\therefore \angle BAC+\angle BCA={90}^{\circ }...\left(i\right)$

In $△ADC,$ we have

$\angle ADC={90}^{\circ }$ [$△ADC$ is a right angled triangle at $D$]

$\angle ADC+\angle DAC+\angle DCA={180}^{\circ }$ [Angle sum property of a triangle]

$\therefore \angle DAC+\angle DCA={90}^{\circ }...\left(ii\right)$

Adding $\left(i\right)$ and $\left(ii\right)$, we get

$\angle BAC+\angle BCA+\angle DAC+\angle DCA={90}^{\circ }+{90}^{\circ }$

$\Rightarrow (\angle BAC+\angle DAC)+(\angle BCA+\angle DCA)={180}^{\circ }$

$\Rightarrow \angle BAD+\angle BCD={180}^{\circ }$

$\therefore \angle ABC+\angle ADC={180}^{\circ }$

If opposite angles of a quadrilateral are supplementary, then it is a cyclic quadrilateral.

$\therefore \angle CAD=\angle CBD$ ($\because$ Angles in the same segment).