Given that, in ,

∠B = 2∠C and,

D is a mid-point on BC such that AD bisects ∠BAC and AB = CD.

We have to prove that, ∠BAC = 72^o

Now draw the angular bisector of ∠ABC, which meets AC in P

Join PD

Let, ∠ACB = y

∠B = ∠ABC = 2∠C = 2y

Also, ∠BAD = ∠DAC = x

∠BAC = 2x (Therefore, AD is the bisector of ∠BAC)

Now, in

∠CBP = y (Therefore, BP is the bisector of ∠ABC)

∠PCB = y

∠CBP = ∠PCB = y

Therefore, PC = BP

Consider , we have

∠ABP = ∠DCP = y

AB = DC (Given)

PC = BP (From above)

So, by SAS theorem, we have

Now,

∠BAP = ∠CDP

And, AP = DP (By c.p.c.t)

∠BAP = ∠CDP = 2x

Now, in

∠ABD + ∠BAD + ∠ADB = 180^o

∠ADB + ∠ADC = 180^o (Straight angle)

2x + 2y + y = 180^o (Therefore, ∠A = 2x, ∠B = 2y, ∠C = y)

2y + 3y = 180^o (Therefore, x = y)

5y = 180^o

y =

y = 36^o

Therefore, x = y = 36^o

Now,

∠A = ∠BAC = 2x = 2 * 36^o = 72^o

Therefore, ∠BAC = 72^o

Hence, proved