Question

In a parallelogram ABCD, the bisectors of angles at B and C intersect each other at point E. Prove that angle BEC is equal to a right angle.

Answer 1

$GivenABCDisaparallelogram.$

$\angle B\&\angle Careadjacentanglesintheparallelogram$

$so,\angle B+\angle C={180}^0\Rightarrow \angle B={180}^0-\angle C---\left(1\right)$

$GiventhatEB\&ECareanglebisectorsfor\angle B\&\angle Crespectively.$

$So,\angle EBC=\frac{1}{2}\angle B\&\angle ECB=\frac{1}{2}\angle C$

$Nowin△BEC,\angle EBC+\angle ECB+\angle CEB={180}^0$

$\Rightarrow \frac{1}{2}\angle B+\frac{1}{2}\angle C+\angle CEB=180$

$\Rightarrow \angle CEB+\frac{1}{2}(\angle B+\angle C)=180\Rightarrow \angle CEB+\frac{1}{2}\times 180={180}^0\left\{from\right(1\left)\right\}$

$\Rightarrow \angle CEB+{90}^0={180}^0$

$\Rightarrow \angle CEB={90}^0$

$\Rightarrow \angle CEBisarightangle.$