Question

In a quadrilateral $ABCD$, the bisector of $\angle C$ and $\angle D$ intersect at $O$.
Prove that $\angle COD=\frac{1}{2}(\angle A+\angle B)$

Answer 1

In given figure $ABCD$ is a quadrilateral.

$\Rightarrow$ CO and DO are bisector of $\angle C$ and $\angle D$

In $△COD$,

$\Rightarrow$ $\angle COD+\angle 1+\angle 2={180}^{\circ }$ [Sum of all interior angles of triangle is ${180}^{c}irc$]

$\Rightarrow$ $\angle COD={180}^{\circ }-(\angle 1+\angle 2)$

$\Rightarrow$ $\angle COD={180}^{\circ }-(\frac{1}{2}\angle C+\frac{1}{2}\angle D)$

$\Rightarrow$ $\angle COD={180}^{\circ }-\frac{1}{2}(\angle C+\angle D)$

$\Rightarrow$ $\angle COD={180}^{\circ }-\frac{1}{2}[{360}^{\circ }-(\angle A+\angle B\left)\right]$

$\Rightarrow$ $\angle COD={180}^{\circ }-{180}^{\circ }+\frac{1}{2}(\angle A+\angle B)$

$\therefore$ $\angle COD=\frac{1}{2}(\angle A+\angle B)$