In parallelogram ABCD, E is the mid-point of side AB and CE bisects angle BCD. Prove that :

(i) AE = AD

(ii) DE bisects angle ADC

(iii) angle DEC is a right angle

(i) In parallelogram ABCD,

AB || DC (Opposite sides are parallel)

CE is the transversal.

∴ ∠CED = ∠BCE (Alternate angles are equal)

∴ BC = EB (In triangle, side opposite to equal angles are equal) ……….(1)

From figure,

⇒ BC = AD (Opposite sides of parallelogram are equal) ……..(2)

Given,

⇒ AE = EB (As, E is the mid-point of AB) ……..(3)

From equation (1), (2) and (3), we get :

⇒ AE = AD.

Hence, proved that AE = AD.

(ii) In △ AED,

⇒ AD = AE (Proved above)

⇒ ∠AED = ∠ADE (Angles opposite to equal sides are equal.) …….(4)

From figure,

⇒ ∠EDC = ∠AED (Alternate angles are equal.) ……….(5)

From equations (4) and (5), we get :

⇒ ∠ADE = ∠EDC.

Hence, proved that DE bisects angle ADC.

(iii) We know that,

Sum of consecutive angles in a parallelogram equal to 180°.

∴ ∠D + ∠C = 180°

⇒ $\dfrac{∠D + ∠C}{2} = \dfrac{180°}{2}$

⇒ $\dfrac{∠D}{2} + \dfrac{∠C}{2}$ = 90°

⇒ ∠EDC + ∠ECD = 90° [As, DE and CE are bisectors of angle D and C] ………..(1)

In △ DEC,

By angle sum property of triangle,

⇒ ∠EDC + ∠ECD + ∠DEC = 180°

⇒ 90° + ∠DEC = 180°

⇒ ∠DEC = 180° - 90° = 90°.

Hence, proved that DEC is a right angle.