ABCD is a trapezium in which BC‖AD and AC=CD. If ∠ABC=40° and ∠BAC=120°, then what is the measure of ∠ACD (in degree)?

Correct option is C

​Given:

Trapezium ABCD with B$$C \parallel A$$​D.

AC=CD so $$\triangle A$$​CD is isosceles with base AD.

​$$\angle ABC=40^\circ,\ \angle BAC=120^\circ.$$​​

Find $$\angle ACD.$$​​

Formula Used:

Triangle angle sum: $$\angle A+\angle B+\angle C=180^\circ.$$​​

In isosceles $$\triangle ACD$$ with AC = CD: base angles $$\angle CAD=\angle ADC.$$​​

If two lines are parallel, corresponding/alternate interior angles with a transversal are equal.

Solution:

In $$\triangle ABC:$$​​
​$$\angle BCA=180^\circ-\angle BAC-\angle ABC=180^\circ-120^\circ-40^\circ=20^\circ.$$​​

Since B$$C \parallel AD,$$​ the angle that CA makes with AD equals the angle it makes with BC. Hence
​$$\angle CAD=\angle ACB=20^\circ.$$​​

In $$\triangle ACD, AC=CD \implies \angle CAD=\angle ADC=20^\circ$$​​

Therefore
​$$\angle ACD=180^\circ-20^\circ-20^\circ=140^\circ.$$​​