Question

In a parallelogram ABCD, the bisector of $\angle ABC$ intersects AD at E. If ED = 5, BE = 6, CE = 6, find AB?

• A. 8
• B. 4
• C. 9
• D. 6

Answer 1

The correct option is B 4


In $\Delta CBE$, using sine rule $=\frac{5+x}{sin({180}^{\circ }-2\theta )}=\frac{6}{sin\theta }$
$\Rightarrow cos\theta =\frac{5+x}{12}$
In $\Delta ABE,cos\theta =\frac{x^2+36-x^2}{12x}$
$\Rightarrow \frac{36}{12x}=\frac{5+x}{12}\Rightarrow x^2+5x-36=0$
$\Rightarrow (x+9)(x-4)=0$
$\Rightarrow x=4$