Let, angle ∠CEB=α, ∠BED=...
Let, angle ∠AEB=α and ∠BED=β
In the triangle, ECD, ∠ECD=90°-α
In the triangle ABC,
∠EAB=Exterior ∠ECB-interior ∠ABC=90°-α-θ= 90°-(α+θ)
In the triangle AEB,
AB/Sin ∠AEB=EB/Sin ∠AEB
AB= EB. Sin ∠AEB/Sin ∠AEB
AB= EB. Sin (α-β)/ Sin {90°-(α+θ)}
AB= EB. Sin (α-β)/ Cos(α+θ)}
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