Question

In an isosceles triangle $ABC,AC=BC,\angle BAC$ is bisected by AD where D lies on BC. It is found that AD = AB. Then $\angle ACB$ equals .

• A. ${72}^o$
• B. ${54}^o$
• C. ${36}^o$
• D. ${60}^o$

Answer 1

The correct option is C ${36}^o$

In $△ABC,AC=BC=>\angle BAC=\angle ABC=x\left(say\right)$

In $△ABD,AB=AD=>\angle ABD=\angle ADB=y\left(say\right)$

But $\angle ABD=\angle ABC=>x=y$

Also $\angle BAD=180°-2x(in△ABD,sumof\angle BAD,\angle ABD,\angle ADBis180°)$

But $AD$ bisects $\angle BAC$ into $\angle BAD$ and $\angle DAC$

$\therefore \angle BAD=\angle DAC=\frac{x}{2}=>180°-2x=\frac{x}{2}=>180°=\frac{5x}{2}=>x=72°\therefore \angle ACB=180°-144°=36°$