Question

In a quadrilateral ABCD, CO and DO are the bisector of $\angle Cand\angle D$ respectively.

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

Answer 1

In quad. ABCD,

CO and DO are the bisectors of $\angle Cand\angle D$ respectively.

In $\Delta COD,\angle DCO+\angle CDO+\angle COD={180}^{\circ }\left(\text{Sum of angles of a triangle}\right)\Rightarrow \frac{1}{2}\angle C+\frac{1}{2}\angle D+\angle COD={180}^{\circ }\Rightarrow \angle COD={180}^{\circ }-(\frac{1}{2}\angle C+\frac{1}{2}\angle D)={180}^{\circ }-\frac{1}{2}(\angle C+\angle D)But\angle A+\angle B+\angle C+\angle D={360}^{\circ }\Rightarrow \angle C+\angle D={360}^{\circ }-(\angle A+\angle B)\therefore \angle COD={180}^{\circ }-\frac{1}{2}[{360}^{\circ }-\angle A-\angle B]={180}^{\circ }-{180}^{\circ }+\frac{1}{2}(\angle A+\angle B)=\frac{1}{2}(\angle A+\angle B)$

Hence proved