Question

In the given quadrilateral ABCD, the line segments bisecting $\angle C$ and $\angle D$ meet at O. Then the correct statement is:

• A. $\angle A+\angle B=\angle COD$
• B. $\angle A+\angle B=2\angle COD$
• C. $\angle A+\angle B=3\angle COD$
• D. $\angle A+\angle B=4\angle COD$

Answer 1

The correct option is B $\angle A+\angle B=2\angle COD$

In the given quadrilateral ABCD,

$\angle 1=\frac{1}{2}\angle C$, $\angle 2=\frac{1}{2}\angle D$
in $△COD,$
$\angle 1+\angle 2+\angle COD={180}^{\circ }$
$\angle COD={180}^{\circ }-(\angle 1+\angle 2)$ ...(i)
Also,
$\angle A+\angle B+\angle C+\angle D={360}^{\circ }$
$\angle A+\angle B+2(\angle 1+\angle 2)={360}^{\circ }$
$\angle A+\angle B={360}^{\circ }-2(\angle 1+\angle 2)$
$\angle A+\angle B=2[{180}^{\circ }-(\angle 1+\angle 2\left)\right]$
Now from (i),
$\angle A+\angle B=2\angle COD$