ABCD is a trapezium in which BC || AD and AC = CD. If ∠ABC = 54° and ∠BAC = 60°, then what is the measure of ∠ACD (in degree)?

Correct option is B

Given:

ABCD is a trapezium with $$BC \parallel AD$$​​

AC = CD, $$\angle ABC = 54^\circ, \angle BAC = 60^\circ$$​​

Find  $$\angle ACD$$​​

Formula Used:

In triangle:  $$\angle A + \angle B + \angle C = 180^\circ$$​​

Corresponding/alternate angles when transversal cuts parallel lines are equal.

If sides are equal, angles opposite them are equal.

Solution: 

In triangle ABC: 

​$$\angle BCA = 180^\circ - (54^\circ + 60^\circ) = 66^\circ$$​​

Since BC $$\parallel$$​ AD and AC is a transversal:

​$$\angle CAD = \angle BCA = 66^\circ$$​​

Given AC = CD, so

​$$\angle CDA = \angle CAD = 66^\circ$$​​

Let $$\angle ACD$$​ = x

In triangle ACD:

​$$x + 66^\circ + 66^\circ = 180^\circ\\\ \\x + 132^\circ = 180^\circ \\\ \\x = 48^\circ$$​​