ABCD is a trapezium in which BC || AD and AC = CD. If ∠ABC = 19° and ∠BAC = 137°, then what is the measure of ∠ACD (in degree)?

Correct option is C

Given:

ABCD is a trapezium

BC $$\parallel$$ AD, AC = CD

​$$\angle ABC = 19^\circ \, \angle BAC = 137^\circ$$​​

Find: $$\angle$$​ ACD

Concept Used:

In triangle ABC:

​$$\angle ABC + \angle BAC + \angle ACB = 180^\circ$$​​

Solution: 

In triangle ABC

​$$\angle ABC = 19^\circ,\quad \angle BAC = 137^\circ$$​​

So, 

​$$\angle ACB = 180^\circ - (19^\circ + 137^\circ) \\ \ \\ \angle ACB = 24^\circ$$​​

Using the parallel-line condition

BC $$\parallel$$​ AD So, 

​$$\angle ACB = \angle CAD = 24\degree$$(alternate angle)​

In triangle ACD

AC = CD => base angles at A and D equal.

​$$\angle CAD = \angle ADC = 24 \degree$$

Now, 

24 + 24 + $$\angle ACD$$​ = 180

$$\angle ACD = 180 - 48 = 132\degree$$​