In the figure, AB || DC and AD || BC. To prove that AB = CD, one must first show that AABC-ACDA.
Which theorem or postulate can be used to justify AABC - ACDA?

The answer is below

Step-by-step explanation:

A triangle is a polygon with three sides and three angles. Types of triangles are isosceles, scalene, equilateral, obtuse, acute and right angled triangles.

Two triangles are said to be congruent if all the corresponding sides and angles of both triangles are equal. Congruent triangles have the same size and shape.

Given that AB || DC and AD || BC, therefore:

∠DCA = ∠BAC (alternate angles are equal)

∠DAC = ∠BCA (alternate angles are equal)

AC ≈ AC

Therefore from triangle ABC and triangle CDA, since ∠DCA = ∠BAC, ∠DAC = ∠BCA and AC ≈ AC, hence both triangles are congruent by Ange-side-angle congruence theorem.

Therefore AB = CD