Question

In a quadrilateral ABCD, the bisectors of C and D meet at E. What is the value of $2\left(\angle CED\right)$?

• A. $\angle A+\angle B$
• B. $\angle A-\angle B$
• C. $\angle C+\angle D$
• D. $\angle A-\angle B$

Answer 1

The correct option is A $\angle A+\angle B$

Consider a quadrilateral ABCD.
As the bisectors of C and D meet at E, if $\angle C=x$ and $\angle D=y$, then $\angle ECD=\frac{x}{2}$ and $\angle EDC=\frac{y}{2}$

Consider sum of angles in triangle CED,
$\angle ECD+\angle CED+\angle EDC={180}^0$
$\frac{x}{2}+\angle CED+\frac{y}{2}={180}^0$
$\angle CED=180-\left(\frac{x+y}{2}\right)$
$2\angle CED=360-(x+y)$............(1)
In Quadrilaterl ABCD, $\angle A+\angle B+\angle C+\angle D={360}^0$
If $\angle C,\angle D$ are x and y then,
$\angle A+\angle B=360-(x+y)$...........(2)
Hence, $2\angle CED=\angle A+\angle B$.
So, the correct answer is option (a).