Question

In the given figure ΔABC is an isosceles traingle with AB=AC and $\angle$ABC = ${50}^{\circ }$. Find $\angle$ BEC = \({x}^{0}\, find the value of x.

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Answer 1

AB = AC
$\angle$ACB = $\angle$ABC
$\angle$ACB = ${50}^{\circ }$ [$\angle$ABC = ${50}^{\circ }$]
$\therefore$$\angle$BAC = ${180}^{\circ }$ - ($\angle$ABC + $\angle$ACB)
$\angle$BAC = ${180}^{\circ }$ - (${50}^{\circ }$ + ${50}^{\circ }$) =${80}^{\circ }$
Since $\angle$BAC and $\angle$BDC are angles in the same segment.
$\therefore$ $\angle$BDC =$\angle$BAC [$\angle$BDC = ${80}^{\circ }$]
Now, BDCE is a cyclic quadrilateral.
$\therefore$ $\angle$BDC + $\angle$BEC = ${180}^{\circ }$
${80}^{\circ }$ + $\angle$BEC = ${180}^{\circ }$
$\angle$BEC = ${100}^{\circ }$
Hence, $\angle$BDC = ${80}^{\circ }$and $\angle$BEC = ${100}^{\circ }$