Question

In the figure, $ABC$ is a triangle in which $AD$ bisects $\angle A$, $AC=$ $BC$,$\angle B={72}^{\circ }$ and $CD=1cm$. Length of $AB\left(incm\right)$ is

• A. $1$
• B. $\frac{1}{2}$
• C. $\frac{\sqrt{5}-1}{2}$
• D. $\frac{\sqrt{3}+1}{2}$

Answer 1

The correct option is A $1$

Given $\angle CAD=\angle ACD=36°$

Hence, $CD=AD=1cm$..........[sides opposite to equal angles are equal]

Also $\angle ADB=\angle ABD=72°$

So, $AB=AD=1cm$.......[sides opposite to equal angles are equal]