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 $ABCD$, by angle sum property

$\angle A+\angle B+\angle C+\angle D={360}^o$

$\therefore \angle A+\angle B={360}^o-(\angle C+\angle D)$

$\frac{1}{2}(\angle A+\angle B)={180}^o-\frac{1}{2}(\angle C+\angle D)$ ...... $\left(1\right)$

Now, In $△COD$

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

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

From $\left(1\right)$ & $\left(2\right)$

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

Hence proved.