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
hope it will help you
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Answers & Comments
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
hope it will help you
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