Question

In AABC, 2B = 35%, 2C = 65° and the bisector AD of ZBAC meets BC at D. Then, which of the following is true​

Answer 1

Answer:

It is given in the question that,

In



, we have

∠B = 35o

∠C = 65o

Also the bisector AD of ∠BAC meets at D

∴ ∠A + ∠B + ∠C = 180o

∠A + 35o + 65o = 180o

∠A = 180o – 100o

∠A = 80o

As, AD is the bisector of ∠BAC

∴ ∠BAD = ∠CAD = 40o

In



, we have

∠BAD > ∠ABD

BD > AD

Also, in



∠ACD > ∠CAD

AD > CD

Hence, BD > AD > CD

∴ Option (B) is correct