Question

In a quadrilateral $ABCD$, the line segments bisecting $\angle C$ and $\angle D$ meet at $E$. prove that $\angle A+\angle B=2\angle CED$

Answer 1

In quadrilateral $ABCD,$

$\angle A+\angle B+\angle C+\angle D={360}^o$ --- ( 1 ) [ Sum of angles of quadrilateral is ${360}^o$ ]

In $△CED,$

$\Rightarrow$ $\angle CED+\angle EDC+\angle ECD={180}^o$ [ Sum of angles of triangle is ${180}^o.$ ]

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

$\Rightarrow$ $2\angle CED+\angle D+\angle C={360}^o$ ----- ( 2 )

From ( 1 ) and ( 2 ),

$\Rightarrow$ $2\angle CED+\angle D+\angle C=\angle A+\angle B+\angle C+\angle D$

$\Rightarrow$ $2\angle CED=\angle A+\angle B$

$\Rightarrow$ $\angle A+\angle B=2\angle CED$