# Thread: Modern Geometry: Complete SsA proof.

1. ## Modern Geometry: Complete SsA proof.

Below is the original problem. After doing some online research, I came to the conclusion that an actual "SSA" Congruence does not exist, so I was a little confused with what this problem was asking. Anyway, below the picture are my attempts at the answers for the missing parts:

My work so far:
(a) I don't know how these two angles can be supplementary . I did find a corollary that states, "The sum of the measrues of 2 angels of a triangle is less than 180."

(b) XY > YZ

(c) Contradiction to the given that BC ≥ BA.

* I think I got (b) and (c), but part (a) is still throwing me off.

Thank you for your time, help, suggestions/corrections.

2. This depends crucially on the condition $\displaystyle BC\geqslant BA$. Notice that if the length BC is less than BA then the triangles XYZ and XYZ' in this picture would both have the same SSA data as ABC (where YZ'=YZ=BC), although XYZ' is not congruent to ABC:

$\displaystyle \setlength{\unitlength}{6mm} \begin{picture}(10,10) \put(0.5,2){\line(1,0){9}} \put(0.5,2){\line(3,2){6}} \put(9.5,2){\line(-3,4){3}} \put(3.5,2){\line(3,4){3}} \put(0.4,1.25){$X$} \put(3.4,1.25){$Z'$} \put(9.4,1.25){$Z$} \put(6.4,6.25){$Y$} \end{picture}$

That is a typical SSA situation, where the position of Z is ambiguous.

But if $\displaystyle BC\geqslant BA$ then the position of Z' would be to the left of X and the triangle XYZ would look like this:

$\displaystyle \setlength{\unitlength}{6mm} \begin{picture}(15,10) \put(0.5,2){\line(1,0){12}} \put(0.5,2){\line(6,5){6}} \put(12.5,2){\line(-6,5){6}} \put(4.5,2){\line(2,5){2}} \put(0,1.25){$Z'$} \put(4,1.25){$X$} \put(12,1.25){$Z$} \put(6.25,7.25){$Y$} \end{picture}$

Notice that in the triangle XYZ' the angle at X is now the supplement of what it is in the triangle XYZ. So the triangle XYZ' does not have the same SSA data as ABC in this case.

I think that is why the question asks for an "SsA" proof, with the second s in lower case. It's essential that the side AB should be shorter than BC.