ABC is a triangle in which ∠B = 2∠C. D is a point on BC such that AD bisects ∠BAC and AB = CD.

Given that in ABC, ∠B = 2 ∠C and D is a point on BC such that AD bisectors ∠BAC and AB = CD.

We have to prove that ∠BAC = 72°

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

Join PD

Let C = ∠ACB = y

∠B = ∠ABC = 2∠C = 2y and also

Let ∠BAD = ∠DAC

∠BAC = 2x [AD is the bisector of ∠BAC]

Now, in ΔBPC,

∠CBP = y [BP is the bisector of ∠ABC]

∠PCB = y

∠CBP = ∠PCB = y [PC = BP]

Consider, ΔABP and ΔDCP, we have

ΔABP = ΔDCP = y

AB = DC                            [Given]

And PC = BP                     [From above]

So, by SAS congruence criterion, we have ΔABP ≅ ΔDCP

Now,

∠BAP = ∠CDF and AP = DP [Corresponding parts of congruent triangles are equal]

∠BAP = ∠CDP = 2

Consider, ΔAPD,

We have AP = DP

= ∠ADP = ∠DAP

But ∠DAP = x

∠ADP = ∠DAP = x

Now

In ΔABD.

∠ABD + ∠BAD + ∠ADB = 180

And also ∠ADB + ∠ADC = 180° [Straight angle]

From the above two equations, we get

∠ABD + ∠BAD + ∠ADB = ∠ADB + ∠ADC

2y + x = ∠ADP + ∠PDC

2y + x = x + 2x

2y = 2x

y = x (or) x = y

We know,

Sum of angles in a triangle = 180°

So, In ΔABC,

∠A + ∠B + ∠C =180°

2x + 2y + y = 180° [∠A = 2x, ∠B = 2y, ∠C = y]

2(y) + 3y =180° [x = y]

5y = 180°

y = 36°

Now, ∠A = ∠BAC = 2 × 36° = 72°