Question
In the given figure, side BC of $\triangle\text{ABC}$ is produced to point D such that bisectors of $\angle\text{ACD}$ meet at a point E. If $\angle\text{BAC}=68^\circ,$ find $\angle\text{BEC}.$
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Answer
In the given figure, bisectors of $\angle\text{ABC}$ and $\angle\text{ACD}$ meet at E and $\angle\text{BAC}=68^\circ$ We need to find $\angle\text{BEC}$ 
Here, using the property: an exterior angle of the triangle is equal to the sum of the opposite interior angles. In $\triangle\text{ABC}$ with $\angle\text{ACD}$ as its exterior angle$\text{ext}.\angle\text{ACD}=\angle\text{A}+\angle\text{ABC}\dots(1)$
Similarly, in $\triangle\text{BE}$ with $\angle\text{ECD}$ as its exterior angle$\text{ext}.\angle\text{ECD}=\angle\text{EBC}+\angle\text{BEC}$
$\frac{1}{2}\angle\text{ACD}=\frac{1}{2}\angle\text{ABC}+\angle\text{BEC}$ $($CE and BE are the bisectors of $\angle\text{ACD}$ and $\angle\text{ABC})$
$\angle\text{BEC}=\frac{1}{2}\angle\text{ACD}-\frac{1}{2}\angle\text{ABC}\dots(2)$
Now, multiplying both sides of (1) by $\frac{1}{2}$ We get,$\frac{1}{2}\angle\text{ACD}=\frac{1}{2}\angle\text{A}+\frac{1}{2}\angle\text{ABC}$
$\frac{1}{2}\angle\text{A}=\frac{1}{2}\angle\text{ACD}-\frac{1}{2}\angle\text{ABC}\dots(3)$
From (2) and (3) we get,$\angle\text{BEC}=\frac{1}{2}\angle\text{A}$
$\angle\text{BEC}=\frac{1}{2}(68^\circ)$
$\angle\text{BEC}=34^\circ$
Thus, $\angle\text{BEC}=34^\circ$
Here, using the property: an exterior angle of the triangle is equal to the sum of the opposite interior angles. In $\triangle\text{ABC}$ with $\angle\text{ACD}$ as its exterior angle$\text{ext}.\angle\text{ACD}=\angle\text{A}+\angle\text{ABC}\dots(1)$Similarly, in $\triangle\text{BE}$ with $\angle\text{ECD}$ as its exterior angle$\text{ext}.\angle\text{ECD}=\angle\text{EBC}+\angle\text{BEC}$
$\frac{1}{2}\angle\text{ACD}=\frac{1}{2}\angle\text{ABC}+\angle\text{BEC}$ $($CE and BE are the bisectors of $\angle\text{ACD}$ and $\angle\text{ABC})$
$\angle\text{BEC}=\frac{1}{2}\angle\text{ACD}-\frac{1}{2}\angle\text{ABC}\dots(2)$
Now, multiplying both sides of (1) by $\frac{1}{2}$ We get,$\frac{1}{2}\angle\text{ACD}=\frac{1}{2}\angle\text{A}+\frac{1}{2}\angle\text{ABC}$
$\frac{1}{2}\angle\text{A}=\frac{1}{2}\angle\text{ACD}-\frac{1}{2}\angle\text{ABC}\dots(3)$
From (2) and (3) we get,$\angle\text{BEC}=\frac{1}{2}\angle\text{A}$
$\angle\text{BEC}=\frac{1}{2}(68^\circ)$
$\angle\text{BEC}=34^\circ$
Thus, $\angle\text{BEC}=34^\circ$
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