If in the given figure, ∠ACB + ∠BAC = 80°, ∠BDE = 35°, ∠BCE = 45°, then the marked angle ∠CED is:

Answer: Option C

∠BAC + ∠ACB = ∠CBD
80° = ∠CBD
∠CBD = 80°
Let, GH || BD
∠DEH = 35°
∠CGE = 80°
∠CGE + ∠GCE = ∠CEH
80° + 45° = ∠CEH
∠CEH = 125°
∠DEH + ∠CEH = ∠CED
125° + 35° = ∠CED
∠CED = 160°