Question
In the given figure, $\triangle\text{ABC}$ is equilateral. Find
$i. \angle\text{BDC}$
$ii. \angle\text{BEC}$
$i. \angle\text{BDC}$
$ii. \angle\text{BEC}$
✓
Answer
$i.$ Given: $\triangle\text{ABC}$ is an equilateral triangle
i.e., each of its angle $= 60^\circ$
$\Rightarrow\ \angle\text{BAC}=\angle\text{ABC}=\angle\text{ACB}=60^\circ$
Angles in the same segment of a circle are equal.
i.e.,$\angle\text{BDC}=\angle\text{BAC}=60^\circ$
$\therefore\ \angle\text{BDC}=60^\circ$
$ii.$ The opposite angles of a cyclic quadrilateral are supplementary.
Then in cyclic quadrilateral $\text{ABEC}$, we have:
$\angle\text{BAC}+\angle\text{BEC}=180^\circ$
$\Rightarrow\ 60^\circ+\angle\text{BEC}=180^\circ$
$\Rightarrow\ \angle\text{BEC}=(180^\circ-60^\circ)=120^\circ$
$\therefore\ \angle\text{BDC}=60^\circ$ and $\angle\text{BEC}=120^\circ$
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