Justify the last two steps of the proof.
Given: ABCD is a parallelogram.
Prove: ΔABC is congruent to ΔCDA
The figure shows parallelogram A B C D with diagonal A C.
ABDC is a parallelogram.
AB is congruent to DC and BC is congruent to DA
AC is congruent to CA
ΔABC is congruent to ΔCDA
Given
Opposite sides of a parallelogram are congruent.
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(1 point)
In Step 4, it is stated that AC is congruent to CA. This is a redundant statement because it is saying that a segment is congruent to itself, which is always true due to the reflexive property of congruence.
In Step 5, it is stated that ΔABC is congruent to ΔCDA. This is justified by the SAS (Side-Angle-Side) Congruence Postulate. We know that AB is congruent to DC and BC is congruent to DA from Step 2. Additionally, we know that AC is congruent to AC (from Step 4, although this step is redundant). Therefore, we have the shared side AC and the two pairs of congruent sides, AB congruent to DC and BC congruent to DA, which satisfies the SAS criterion for congruence of triangles.