Question

$\text{In parallelogram ABCD, E is the mid-point of side AB and CE bisects angle BCD.}\text{Prove that :}\left(i\right)AE=AD.\left(ii\right)\text{DE bisects and}\angle ADCand\left(iii\right)\text{Angle DEC is a right angle.}$

Answer 1

$\text{Given :}||\text{gm ABCD in which E is mid-point of AB and CE bisects}\angle BCD.\text{To Prove : (i) AE = AD}\left(ii\right)\text{DE bisects}\angle ADC\left(iii\right)\angle DEC={90}^{\circ }\text{Const. Join DE}\text{Proof :}\left(i\right)AB||CD\text{(Given)}\text{and CE bisects it.}\therefore \angle 1=\angle 3\left(\text{alternte}\angle s\right)...\left(i\right)But\angle 1=\angle 2\text{(Given)}...\left(ii\right)\text{From (i)}\&\left(ii\right)\angle 2=\angle 3\therefore BC=BE\text{(sides opp. to equal angles)}ButBC=AD\left(\text{opp. sides of}||gm\right)andBE=AE\text{(Given)}AD=AE\therefore \angle 4=\angle 5\left(\angle s\text{opp. to equal sides}\right)But\angle 5=\angle 6\left(\text{alternate}\angle s\right)\Rightarrow \angle 4=\angle 6\therefore DEbisects\angle ADC.NowAD||BC\Rightarrow \angle 4=\angle 6\therefore DEbisects\angle ADC.NowAD||BC\Rightarrow \angle D+\angle C={180}^{\circ }\therefore 2\angle 6+2\angle 1={180}^{\circ }\because \text{DE and Ce are bisectors.}\angle 6+\angle 1=\frac{{180}^{\circ }}{2}\angle 6+\angle 1={90}^{\circ }But\angle DEC+\angle 6+\angle 1={180}^{\circ }\therefore \angle DEC+{90}^{\circ }={180}^{\circ }\angle DEC={180}^{\circ }-{90}^{\circ }\angle DEC={90}^{\circ }\text{Hence the result.}$