Justify the last two steps of the proof.%0D%0A%0D%0AGiven: ABCD is a parallelogram.%0D%0AProve: ΔABC ΔCDA%0D%0A%0D%0A%0D%0A%0D%0AABDC is a parallelogram.%0D%0AAB DC and BC DA%0D%0AAC CA%0D%0AΔABC ΔCDA%0D%0AGiven%0D%0AOpposite sides of a parallelogram are congruent.%0D%0A_____________________?%0D%0A_____________________?
The last two steps of the proof are justified as follows:
1. Opposite sides of a parallelogram are congruent: This property states that if ABCD is a parallelogram, then AB is congruent to DC and BC is congruent to DA.
2. If two sides of a triangle are congruent to two sides of another triangle, and the included angles are congruent, then the two triangles are congruent by SSS (side-side-side) congruence: In this case, we have shown that AB is congruent to DC (from opposite sides of a parallelogram property) and BC is congruent to DA (from the same property). We also know that angle AC is congruent to angle CA (given). Therefore, by the SSS congruence property, we can conclude that triangle ABC is congruent to triangle CDA.