Question

Suppose $ABC$ is an equilateral triangle. Its base $BC$ is produced to $D$ such that $BC=CD$. Calculate (i) $\angle ACD$ (ii) $\angle ADC$.

Answer 1

Given: $ABC$ is an equilateral triangle,

$\therefore AB=BC=CA$,

and also $BC=CD$.

Proof: In $△ABC$,

$\angle A=\angle B=\angle C={60}^{\circ }$ (Equilateral Triangle)

$\Rightarrow \angle ACB={60}^{\circ }$

Now, $\angle ACB+\angle ACD={180}^{\circ }$ ($\because$ linear pair)

$\Rightarrow {60}^{\circ }+\angle ACD={180}^{\circ }$

$\Rightarrow \angle ACD={180}^{\circ }-{60}^{\circ }$

$\Rightarrow \angle ACD={120}^{\circ }$

In $△ACD$ , $\angle ADC=\angle CAD$ $(\because AC=CD)$

$\Rightarrow \angle ADC=\angle CAD=\frac{{180}^{\circ }-\angle ACD}{2}$

$\Rightarrow$ $\angle ADC=\frac{{180}^{\circ }-\angle ACD}{2}$

$\Rightarrow$ $\angle ADC=\frac{{180}^{\circ }-{120}^{\circ }}{2}$

$\Rightarrow$ $\angle ADC=\frac{{60}^{\circ }}{2}$

$\Rightarrow$$\angle ADC$ = ${30}^{\circ }$

Hence, the required values are, $\angle ACD={120}^{\circ }$ and $\angle ADC$ = ${30}^{\circ }$.