Which would prove that triangle ABC is congruent to triangle XYZ? select 2 that apply

A and C

Step-by-step explanation:

You put the same sides as a fraction equal to the other sides as fractions. For example, side AB is the same as side XY. You would write it like this: AB/XY and do the rest. It's the same for angles.

The answer is the first option and the second option

,

Step-by-step explanation:

The first option:

The ratios must be the same between all sides to prove that the sides are similar. If the sides are proportional then, the sides are equal. The triangles are already in the same position, so to get the proper ratios, the corresponding sides need to be set up in a equation as fraction divided by their counterparts (corresponding sides). This is present, so the sides are equal.

The second option:

This option is also true because they are the proper fractions that need to equal each other, but also the angle is corresponding to its counterpart.

given that triangle abc is transformed two times to get triangle

a"b"c"

first abc is transformed in to a'b'c' as follows:

the coordinates of a are (1,-1) and transformed into a'(-1,1)

similarly b (4,-2) became b'(-4,2) and

c(7,-2) became c'(-7,2)

i.e. (x,y) becomes (-x,-y)

this is nothing but reflection about a point here origin.

thus first transformation is reflection on the origin.

next is exactly shifting 3 units down

source:

step-by-step explanation:

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