KVPY 2020-21 SA Exam Previous Papers

∵ ∠DAB, ∠ABC and ∠BCD are in A.P.
∴ Let ∠DAB = θ – α, ∠ABC = θ and ∠BCD = θ + α
∴ Median of ∠DAB, ∠ABC and ∠BCD = θ
From point E all the vertices are at equal distance.
∴ quadrilateral is cyclic.
and ∠ADC = 2π – (θ – α + θ + θ + α) = 2π –3θ
and ∠ADC + ∠ABC = π
⇒ 2π – 3θ + θ = π
⇒θ=

π

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