Justify the last two steps of
the proof.
Given: ABCD is a
parallelogram.
Prove: ΔABC ΔCDA
1. ABDC is a
parallelogram.
2. AB DC and
BC DA
3. AC CA
4. ΔABC
ΔCDA
1. Given
2. Opposite sides of a
parallelogram are
congruent.
3. _____________________?
4. _____________________?
(1 point)
Transitive Property of
; SAS
Reflexive Property of
; SAS
Transitive Property of
; SSS
Reflexive Property of
; SSS
The third step can be justified using the Reflexive Property of Congruence. This property states that any segment is congruent to itself. In this case, segment AC is congruent to itself, therefore AC is congruent to CA.
The fourth step can be justified using the SAS (Side-Angle-Side) congruence postulate. This postulate states that if two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle, then the triangles are congruent. In this case, we know that AB is congruent to CD, BC is congruent to DA, and AC is congruent to CA. Therefore, by SAS, triangle ABC is congruent to triangle CDA.
Which question tho!?!?
Apologies for the confusion. The justification should be as follows:
3. Reflexive Property of Congruence (AC is congruent to itself)
4. SAS (Side-Angle-Side) congruence postulate