Given: CB = CD and ∠CBD = 35°

Consider ΔBCD

Here,

CB = CD (given)

∠CBD = ∠CDB = 35° (In a triangle, angles opposite to equal sides are equal)

By angle sum property

∠BCD + ∠CBD + ∠CDB = 180°

∠BCD + 35° + 35° = 180°

∠BCD = 180° – 35° – 35° = 110°

We know that,

In a cyclic quadrilateral opposite angles are supplementary

∴ ∠BCD + ∠BAD = 180°

110° + ∠BAD = 180°

∠BAD = 180° – 110° = 70°

∴ ∠BAD = 70°