Question

$\square \mathrm{ABCD}$ is such a quadrilateral that $\mathrm{A}$ is the centre of the circle passing through $\mathrm{B},\mathrm{C}$ and $\mathrm{D}$. Prove that $\angle \mathrm{CBD}+\angle \mathrm{CDB}=\frac{1}{2}\angle \mathrm{BAD}$

Answer 1

Step:1 Construction:
Join $\mathrm{CA}$ and $\mathrm{BD}$.

Step2: Proving that $\angle \mathrm{CBD}+\angle \mathrm{CDB}=\frac{1}{2}\angle \mathrm{BAD}$

We know that,

In a circle, the angle subtended by an arc at the centre is twice the angle subtended by it at any other point in the remaining part of the circle

The arc $\mathrm{DC}$ subtends $\angle \mathrm{DAC}$ at the centre and$\angle \mathrm{CBD}$at point $\mathrm{B}$ in the remaining part of the circle,

So,

$\angle \mathrm{DAC}=2\angle \mathrm{CBD}$ ——($1$)

The arc $\mathrm{BC}$ subtends $\angle \mathrm{CAB}$ at the center and $\angle \mathrm{CDB}$ at point $\mathrm{D}$ in the remaining part of the circle,

So,

$\angle \mathrm{CAB}=2\angle \mathrm{CDB}$ _______($2$)

Add equations ($1$) and ($2$),

$\begin{array}{rcl}\angle \mathrm{DAC}+\angle \mathrm{CAB}& =& 2\angle \mathrm{CDB}+2\angle \mathrm{CBD}\\ & \Rightarrow & \angle \mathrm{BAD}=2(\angle \mathrm{CDB}+\angle \mathrm{CBD})\\ & \Rightarrow & (\angle \mathrm{CDB}+\angle \mathrm{CBD})=\frac{1}{2}(\angle \mathrm{BAD})\end{array}$

Hence we prove that $\angle \mathrm{CBD}+\angle \mathrm{CDB}=\frac{1}{2}\angle \mathrm{BAD}$