Question

15. In parallelogram ABCD, E is the mid-point
of side AB and CE bisects angle BCD. Prove
that:
(i) AE = AD
​

Answer 1

Answer:

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

Given : ||gm 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)

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