Question

Figure ABCD is a parallelogram.

Which sequence could be used to prove that AD = BC?

First prove ABC is congruent to CDA, and then state AD and BC are corresponding sides of the triangles.

First prove ABC is similar to CDA, and then state AD and BC are opposite sides of the parallelograms.

First prove ABCD is congruent to CDAB, and then state AD and BC are corresponding sides of two parallelograms.

First prove ABCD is similar to CDAB, and then state AD and BC are opposite sides of the parallelograms.

HELP QUICKLY PLEASE!!!!!!!!

Answer 1

Answer: First prove ABC is congruent to CDA, and then state AD and BC are corresponding sides of the triangles.

Explanation:

In the given picture, we have a parallelogram ABCD with diagonal AC..

AB║ CD, AD║BC

Now, By its diagonal AC it is divided in two triangles Δ ABC and Δ ADC

∠ACB=∠DAC and ∠CAB=∠ACD [alternate interior angles]

AC=AC {Reflexive property}

∴ By ASA postulate of congruence ,

Δ ABC ≅ Δ ADC

⇒ AD = BC [corresponding sides of the congruent triangles are congruent]

Hence, The sequence could be used to prove that AD = BC is

"First prove ABC is congruent to CDA, and then state AD and BC are corresponding sides of the triangles. "