ASAP Which condition does not prove that two triangles are congruent? A. ASA ≅ ASA B. SAS ≅ SAS C. SSA ≅ SSA D. SSS ≅ SSS

The answer is C. SSA ≅ SSA.

Step-by-step explanation:

To check for similar triangles, SSA congruence would not work because the other side can be any length. Also, there is not an SSA postulate because this theorem by itself cannot prove congruence.

The other three properties do work because they show congruence unlike the other congruent factors.

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.