Question

In parallelogram $ABCD,E$ is a mid-point of side $AB$ and $CE$ bisect angle $BCD.$ Prove that
Angle $DEC$ is a right angle.

Answer 1

Given : ||gm $ABCD$ in which $E$ is mid-point of $AB$ and $CE$ bisects $\angle BCD.$
To prove: (i) $\angle DEC={90}^o$
Const. join $DE$
Proof: (i) $AB\parallel CD$ (Given)
And $CE$ bisect it.
$\angle 1=\angle 3$ (Alternate $\angle S$).....(i)
But $\angle 1=\angle 2$ (Given)......(ii)
From (i)&(ii)
$\angle 2=\angle 3$
$BC=BE$ (Sides opp. to equal angles)
But $BC=AD$ (opp. sides of ||gm)
And $BE=AE$ (Given)
$AD=AE$
$\angle 4=\angle 5$ ($\angle S$ opp. to equal sides)
But $\angle 5=\angle 6$ (Alternate $\angle S$)
$\Rightarrow \angle 4=\angle 6$
$DE$ bisects $\angle ADC.$
Now $AD//BC$
$\Rightarrow \angle D+\angle C={180}^o$
$2\angle 6+2\angle 1={180}^o$
$DE$ and $CE$ are bisectors.
$\angle 6+\angle 1=\frac{{180}^0}{2}$
$\angle 6+\angle 1={90}^0$
But $\angle DEC+\angle 6+\angle 1={180}^o$
$\angle DEC+{90}^o={180}^o$
$\angle DEC={180}^o-{90}^o$
$\angle DEC={90}^o$
Hence the result.