Question

In the given figure, $\Delta ABC$ is an equilateral triangle and $\Delta BDC$ is an isosceles triangle. Find $\angle ACD$.

Answer 1

​Since, $\Delta ABC$ is an equilateral triangle, therefore, each angle of the triangle $\Delta ABC$ is ${60}^{\circ }$.

Since, $\Delta BDC$ is an isosceles triangle, therefore, $BD=DC$,

$\therefore \angle DBC=\angle BCD=x$ (say)

Using, Angle Sum Property in $\Delta BDC$,

$\Rightarrow \angle BCD+\angle BCD+\angle CBD={180}^{\circ }$

$\Rightarrow x+x+{90}^{\circ }={180}^{\circ }$

$\Rightarrow 2x={90}^{\circ }$

$\Rightarrow x={45}^{\circ }$

Therefore,

$\angle ACD=\angle ACB+\angle DCB={45}^{\circ }+{60}^{\circ }={105}^{\circ }$