Question

Question 11
ABC and ADC are two right triangles with common hypotenuse AC. Prove that $\angle CAD=\angle CBD$.

Answer 1

In $\Delta ABC$,
$\angle ABC+\angle BCA+\angle CAB={180}^{\circ }$ (Angle sum property of a triangle)
$\angle {90}^{\circ }+\angle BCA+\angle CAB={180}^{\circ }$
$\therefore \angle BCA+\angle CAB={90}^{\circ }...\left(1\right)$

In $\Delta ADC$,
$\angle CDA+\angle ACD+\angle DAC={180}^{\circ }$ (Angle sum property of a triangle)
$\angle {90}^{\circ }+\angle ACD+\angle DAC={180}^{\circ }$
$\therefore \angle ACD+\angle DAC={90}^{\circ }...\left(2\right)$

Adding equations (1) and (2), we obtain
$\angle BCA+\angle CAB+\angle ACD+\angle DAC={180}^{\circ }$
$(\angle BCA+\angle ACD)+(\angle CAB+\angle DAC)={180}^{\circ }$
$\angle BCD+\angle DAB={180}^{\circ }...\left(3\right)$
However, it is given that $\angle B+\angle D={90}^{\circ }+{90}^{\circ }={180}^{\circ }...\left(4\right)$
From equations (3) and (4), it can be observed that the sum of the measures of opposite angles of quadrilateral ABCD is ${180}^{\circ }$.
Therefore, it is a cyclic quadrilateral.

Consider chord CD.



$\angle CAD=\angle CBD$ (Angles in the same segment)
Hence proved.