Question

In parallelogram $ABCD,E$ is the mid-point of side $AB$ and $CE$ bisects angle $BCD$. Prove that$:$
$\left(I\right)AE=AD$
$\left(II\right)DE$ bisects $\angle ADC$ and
$\left(III\right)\angle DEC$ is a right angle.

Answer 1

In $△CBE,\angle ECB+\angle CBE+\angle BEC={180}^{\circ }$

$⟹\frac{C}{2}+{180}^{\circ }-C+\alpha ={180}^{\circ }⟹\alpha =\frac{C}{2}$

$⟹△CBE$ is an isosceles triangle $⟹b=\frac{a}{2}$

$⟹AE=AD$

$△AED$ is also an isosceles triangle $(\because AE=AD)$

$⟹\angle AED=\angle ADE=\theta$

$\angle AED+\angle ADE+\angle DAE={180}^{\circ }$

$\theta +\theta +C={180}^{\circ }⟹\theta ={90}^{\circ }-\frac{C}{2}$

$\angle EDC={180}^{\circ }-C-\angle ADE={90}^{\circ }-\frac{C}{2}=\angle ADE$

$⟹DE$ bisects $\angle ADC$

From figure ,$\theta +\alpha +\angle DEC={180}^{\circ }$

$⟹{90}^{\circ }-\frac{C}{2}+\frac{C}{2}+\angle DEC={180}^{\circ }⟹\angle DEC={90}^{\circ }$