Let’s write the measurement of ∠ABC, ∠ACB and ∠BAC if AB = AC

Given ΔABC, ∠ACD = 112° and AB = AC

To find the measurement of ∠ABC, ∠ACB and ∠BAC

In the given figure, BCD is a straight line, so

∠BCD = 180°

But, ∠BCD = ∠ACB + ∠ACD

∴ ∠ACB + ∠ACD = 180°

But given ∠ACD = 112°, so the above equation becomes,

⇒ ∠ACB + 112° = 180°

⇒ ∠ACB = 180° - 112°

⇒ ∠ACB = 68°………(i)

Now consider ΔABC, it is given AB = AC

We know angles opposite to equal sides are equal, therefore

⇒∠ACB = ∠ABC

Now equation (i) in above equation we get

∠ABC = 68°……….(ii)

Considering the same triangle, i.e., ΔABC

We know in a triangle the measure of any exterior angle of a triangle is equal to the sum of the measures of its two interior opposite angles.

So in this case ∠ACD is an exterior angle, therefore

∠ACD = ∠ABC + ∠BAC

Substituting the values from given criteria and equation (ii), we get

112° = 68° + ∠BAC

⇒ ∠BAC = 112° - 68°

⇒ ∠BAC = 44°……..(iii)

Hence from equation (i), (ii) and (iii), the measurements are

∠ABC = 68°, ∠ACB = 68° and ∠BAC = 44°