It is given in the question that,

In , we have

∠B = 35^o

∠C = 65^o

Also the bisector AD of ∠BAC meets at D

∴ ∠A + ∠B + ∠C = 180^o

∠A + 35^o + 65^o = 180^o

∠A = 180^o – 100^o

∠A = 80^o

As, AD is the bisector of ∠BAC

∴ ∠BAD = ∠CAD = 40^o

In , we have

∠BAD > ∠ABD

BD > AD

Also, in

∠ACD > ∠CAD

AD > CD

Hence, BD > AD > CD

∴ Option (B) is correct