In the given figure, we find :

1. BD > AB

2. BD < AB

3. BD = AB

4. DC < AB

In △ ABC,

By angle sum property of triangle,

⇒ ∠A + ∠B + ∠C = 180°

⇒ ∠A + 75° + 35° = 180°

⇒ ∠A + 110° = 180°

⇒ ∠A = 180° - 110° = 70°.

From figure,

AD bisects angle A.

∴ ∠BAD = ∠DAC = $\dfrac{∠A}{2} = \dfrac{70°}{2}$ = 35°.

In △ ABD,

By angle sum property of triangle,

⇒ ∠ABD + ∠BAD + ∠ADB = 180°

⇒ 75° + 35° + ∠ADB = 180°

⇒ ∠ADB + 110° = 180°

⇒ ∠ADB = 180° - 110° = 70°.

Since, ∠BAD < ∠ADB

∴ BD < AB (If two angles of a triangle are unequal, the greater angle has the greater side opposite to it.)

Hence, Option 2 is the correct option.