Let z,o={[0] 10, | | | 10, , [9] 10 } be the set of equivalence classes of z under the relation, "congruence mod- ulo 10," endowed with an addition and a multiplication as presented in lectures. calculate (a)11 (b)岡 (c) for each x e z10 for which it exists, the element y e zo such that xy-yx = [1] 10-

D: Triangles ABD and CDB are congruent by the ASA postulate instead of the SAS postulate.

Step-by-step explanation:

I got it right on the test for one, but two, we must look at the two lines before:

1.) Similarly, alternate interior angle ADB is equal to angle CBD because AD and BC are parallel lines.

So we have two angles, and:

2.) DB is equal to DB by the reflexive property.

So, we see that we have two angles, and one line, what is that? That is ASA! Not SAS, now if the proof provided an angle and a line on 1, and then the reflexive property in 2, then SAS would be correct, but the proof called two angles and that side, so they need to say ASA.