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

(i) AE = AD

(ii) DE bisects ∠ADC and

(iii) Angle DEC is a right angle.

(i) Given:

Parallelogram ABCD, E is the mid-point of side AB and CE bisects angle BCD.

To prove:

AE = AD

Proof:

Firstly join DE.

It is given that ∠ ECB = ∠ ECD.

AB is parallel to CD.

⇒ ∠ CEB = ∠ ECD (alternate angles)

So, ∠ CEB = ∠ ECB

BC = BE (sides opposite to equal angles are always equal)

And, we also know BC = AD (sides of parallelogram)

BE = AE (Given)

So, AD = AE (Using above equations)

Hence, AD = AE.

(ii)To prove:

∠ADE = ∠CDE

Proof:

∠ AED = ∠ ADE (angles opposite to equal sides are always equal)

AB is parallel to CD.

⇒ ∠ AED = ∠ EDC (alternate angles)

So, ∠ ADE = ∠ CDE

Hence, DE bisects ∠ ADC.

(iii)To prove:

∠ DEC = 90°

Proof:

AD is parallel to BC.

⇒ ∠D + ∠C = 180°

⇒ 2∠EDC + 2∠DCE = 180°

⇒ 2(∠EDC + ∠DCE) = 180°

⇒ ∠EDC + ∠DCE = $\dfrac{180°}{2}$

⇒ ∠EDC + ∠DCE = 90°

In triangle DEC, sum of all the angles is 180°

⇒ ∠EDC + ∠DCE + ∠DEC = 180°

⇒ 90° + ∠DEC = 180°

⇒ ∠DEC = 180° - 90°

⇒ ∠DEC = 90°

Hence, Angle DEC is a right angle.