Question

In a quadrilateral ABCD, CO and DO are the bisectors of $\angle C$ and $\angle D$ respectively. Prove that $\angle COD=\frac{1}{2}(\angle A+\angle B)$.

Answer 1

In $△COD$,

$\angle COD+\angle 1+\angle 2={180}^o$

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

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

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

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