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in parallelogram ABCD, E is the midpoint of side AD and CE bisects angle BCD prove that

D bisects angle ADC ​

Answer 1

Step-by-step explanation:

In △CBE,∠ECB+∠CBE+∠BEC=180

∘

⟹

2

C

+180

∘

−C+α=180

∘

⟹α=

2

C

⟹△CBE is an isosceles triangle ⟹b=

2

a

⟹AE=AD

△AED is also an isosceles triangle (∵AE=AD)

⟹∠AED=∠ADE=θ

∠AED+∠ADE+∠DAE=180

∘

θ+θ+C=180

∘

⟹θ=90

∘

−

2

C

∠EDC=180

∘

−C−∠ADE=90

∘

−

2

C

=∠ADE

⟹DE bisects ∠ADC

From figure ,θ+α+∠DEC=180

∘

⟹90

∘

−

2

C

+

2

C

+∠DEC=180

∘

⟹∠DEC=90

∘

solution

Answer 2

Answer:

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