Question

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

(i) AE = AD,
(ii) DE bisects and ∠ADC and
(iii) Angle DEC is a right angle.

Answer 1

Given: parallelogram ABCD in which E is mid-point of AB and CE bisects ZBCD.

To Prove : 
(i) AE = AD
(ii) DE bisects ∠ADC
(iii) ∠DEC = 90°

Const. Join DE

Proof : (i) AB || CD (Given)

and CE bisects it.

∠1 = ∠3 (alternate ∠s) ……… (i)

But ∠1 = ∠2 (Given) …………. (ii)

From (i) & (ii)

∠2 = ∠3

BC = BE (sides opp. to equal angles)

But BC = AD (opp. sides of ||gm)

and BE = AE (Given)

AD = AE

∠4 = ∠5 (∠s opp. to equal sides)

But ∠5 = ∠6 (alternate ∠s)

=> ∠4 = ∠6

DE bisects ∠ADC.

Now AD || BC

=> ∠D + ∠C = 180°

2∠6+2∠1 = 180°

DE and CE are bisectors.

∠6 + ∠1 = `180^circ/2`

∠6 + ∠1 = 90°

But ∠DEC + ∠6 + ∠1 = 180°

∠DEC + 90° = 180°

∠DEC = 180° – 90°

∠DEC = 90°

Hence the result.