Question

If the bisectors of two adjacent angles A and B of a quadrilateral ABCD intersect at a point O such that $\angle C+\angle D=k\left(\angle AOB\right)$, then find the value of k.

Answer 1

In quadrilateral ABCD, Bisectors of $\angle A$ and $\angle B$ meet at O, such that $\angle C+\angle D=k\left(\angle AOB\right)$.

$\angle AOB={180}^{\circ }-(\frac{1}{2}\angle A+\frac{1}{2}\angle B)$

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

But $k\left(\angle AOB\right)=\angle C+\angle D$

$\therefore \angle AOB=\frac{1}{k}(\angle C+\angle D)$ ....(ii) From (i) and (ii)

${180}^{\circ }-\frac{1}{2}(\angle A+\angle B)=\frac{1}{k}(\angle C+\angle D)\frac{1}{2}(A+B)+\frac{1}{2}(\angle C+\angle D)-\frac{1}{2}(A+B)=\frac{1}{k}(\angle C+\angle D)(\because \angle A+\angle B+\angle C+\angle D={360}^{\circ })\frac{1}{2}(\angle C+\angle D)=\frac{1}{k}(\angle C+\angle D)\text{Comparing , we get}\frac{1}{k}=\frac{1}{2}\Rightarrow k=2$