Question
In the given figure, If $ABC$ is an equilateral triangle. Find $\angle\text{BDC}$ and $\angle\text{BEC.}$
✓
Answer
Since, $\triangle\text{ABC}$ is an equilateral triangle
Then, $\angle\text{BAC}=60^\circ$
$\therefore\angle\text{BDC}=\angle\text{BAC}=60^\circ$ [Angles in same segment]
Since, quad. $ABEC$ is a cyclic quadrilateral.
Then, $\angle\text{BAC}+\angle\text{BEC}=180^\circ$
$\Rightarrow60^\circ+\angle\text{BEC}=180^\circ$
$\Rightarrow\angle\text{BEC}=180^\circ-60^\circ=120^\circ$
Then, $\angle\text{BAC}=60^\circ$
$\therefore\angle\text{BDC}=\angle\text{BAC}=60^\circ$ [Angles in same segment]
Since, quad. $ABEC$ is a cyclic quadrilateral.
Then, $\angle\text{BAC}+\angle\text{BEC}=180^\circ$
$\Rightarrow60^\circ+\angle\text{BEC}=180^\circ$
$\Rightarrow\angle\text{BEC}=180^\circ-60^\circ=120^\circ$
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