Question

In the given figure ABCD is a parallelogram E is mid-point of AB and CE bisects ZBCD. Prove that DA D AE AD 1) DE bisects ADC 18) DEC = 90 AL A B E​

Answer 1

Answer:

In △CBE,∠ECB+∠CBE+∠BEC=180∘

              ⟹2C+180∘−C+α=180∘⟹α=2C

⟹△CBE is an isosceles triangle ⟹b=2a

⟹AE=AD

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

⟹∠AED=∠ADE=θ

∠AED+∠ADE+∠DAE=180∘

θ+θ+C=180∘⟹θ=90∘−2C

∠EDC=180∘−C−∠ADE=90∘−2C=∠ADE

⟹DE bisects ∠ADC

From figure ,θ+α+∠DEC=180∘

⟹9

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