Question

In rectangle 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

REF.IMAGE

In $\Delta COD$

$\angle COD+\angle 1+\angle 2={180}^0$

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

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

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

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

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