Let G = BE ∩ CD. Let 2α = ∠BAC, β = ∠ABE = ∠EBC,
γ = ∠ACD = ∠DCB, θ = ∠BDC, φ = ∠CEB, and ψ = ∠BGC.
Construct point F on the opposite side of line CD from
vertex B such that ∠CDF = β and ∠DCF = φ.
Therefore, ΔCDF ≅ ΔEBC by angle-side-angle.
This inplies |CF| = |EC| and |DF| = |CB|.
∠BDF = ∠BDC + ∠CDF = θ + β = ψ = φ + γ =
∠FCD + ∠DCB = ∠FCB.
From ΔBGC we have β + γ + ψ = 180° and from
ΔABC we have 2α + 2β + 2γ = 180°
thus ψ = 90° + α > 90°.
Therefore, |DF| = |CB| and |FB| = |BF| imply
ΔBDF ≅ ΔFCB
by obtuse-side-side.
This implies |DB| = |CF|.
|DB| = |CF| = |EC|, |BC| = |CB|, and |CD| = |BE|.
Therefore, ΔDBC ≅ ΔECB by side-side-side.
This implies,
∠B = ∠ABC = ∠DBC = ∠ECB; = ∠ACB = ∠C.
QED
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