Question

15. In parallelogram ABCD, E is the mid-point
of side AB and CE bisects angle BCD. Prove
that:
() AE = AD
(ii) DE bisects and _ADC and
(iii) Angle DEC is a right angle.​

Answer 1

Answer:

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

Step-by-step explanation:

Hy mate that's your answer