Question

in the adjoining figure ,ABCD is a parallelogram, E is the mid-point of AB and CE bisects angle BCD. prove that :
1) AE = AD
2) DE bisects angle ADC
3)angle DEC =90⁰​

Answer 1

Answer:

here u go................

Step-by-step explanation:

ABCD is parallelogram. E is mid-point of AB and CE bisects BCD

AB||CD and CE Traversal

EC Bisects ∠BCD

IN parallelogram. ABCD

∠D+∠C=180

∴

2

1

∠D+

2

1

∠C=

2

1

×180

∠EDC+∠ECD=90

0

△DEC

∠DEC+∠EDC+∠ECD=180

0

∴∠DEC+90

0

=180

0

∴∠DEC=180

0

−90

0

=90

0

Answer 2

Step-by-step explanation:

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(i)AB || CD (Given)

and CE bisects BCD

∠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

(ii) ∠4 = ∠5 (∠s opp. to equal sides)

But ∠5 = ∠6 (alternate ∠s)

=> ∠4 = ∠6

DE bisects ∠ADC.

(iii) Now AD // BC

=> ∠D + ∠C = 180°

2∠6 + 2∠1 = 180°

DE and CE are bisectors.

∠6 + ∠1 = 180°/2

∠6 + ∠1 = 90°

But ∠DEC + ∠6 + ∠1 = 180°

∠DEC + 90° = 180°

∠DEC = 180° – 90°

∠DEC = 90°

Hence the result.