(d) 135°.
Let AC and BD intersect at point K and ∠ABK = β, ∠KBC = α, ∠ADK = γ.
Then, In rt. ΔAKB, ∠BAK = 90° – β
Also, in rt. DAKD, ∠DAK = 90° – γ
∴ AB = BD
∴ ∠BAD = ∠BDA ⇒ (90° – β) + (90° – γ) = γ
⇒ 180° – β – γ = γ
⇒ 2γ = 180° – β ⇒ γ = 90° – \(\frac{β}{2}\). ...(i)
In rt. ΔBKC, ∠C = 90° – α = ∠ABC = α + β (∵AB = AC)
∴ 90° – α = α + β ⇒ 2α = 90° – β ⇒ α = 45° – \(\frac{β}{2}\). ...(ii)
∠C + ∠D = (90° – α) + γ = 90° – (45° – \(\frac{β}{2}\)) + 90° – \(\frac{β}{2}\)
From (i) and (ii)
= 45° + \(\frac{β}{2}\) + 90° – \(\frac{β}{2}\) = 135°.
