Question

The length of $AD$ is

Answer 1

In $△ABC$
$AC=BC$
$\angle ABC=\angle CAB=x$ ...(Isosceles triangle property)
Now, $\angle ABC+\angle CAB=\angle ACD$ ...(Exterior angle property)
$x+x=60$
$x={30}^{\circ }$
Now, In $△ABD$,
$\frac{AD}{AB}=\sin 30$

$\frac{AD}{100}=\frac{1}{2}$

$AD=50$cm