txags911 is right, ioi. There is not enough information to prove they are congruent. There has to be an equal angle between the two equal sides. If you tried to do a proof in Geometry on this, and decided they were equal based on the given information, you'd get it wrong.

Actually, even the way they are drawn, they could be slightly off and not congruent. Let's say you take one of the two angles that look like they are 90 degrees and bend them by 1 degree (or even less). As a result, what appears to be two lines intersecting in the middle will not actually be that. The lines will skew ever so slightly in the center. So one of the lines crossing will not be completely straight. So you've changed angles, left the two congruent sides as being congruent as the picture says, and even maintained the overall look of the picture. The changes are so small they would be barely noticeable. But the triangles are not congruent since one has a 90 degree angle on in and the other has an 89 degree one.

That is true, but you have to remember who he was arguing with, his math teacher.  This means that the only acceptable answer that he can give will need to be proven mathamaticly and not by sight.

There is a 0.000000000000000001% chance that it is perfectly congruent. Maybe the percentage is even smaller than that. However, when the diagram is close enough to perfection, we assume that it is. If we really judged the small imperfections of such diagrams, than Math class would be filled with microscopes.

I understand the point. Just wanted to add that post in.

Wow, I can't believe this thread has got this far :) If the lines are continuous - and I believe AND assume that they are (this is a maths question, not some trick question - it should state that the lines are NOT continuous if they are not!) then the tris are equivalent. PS - on a subnote, a congruency test that is sufficient is for 2 sides to have the same length and ANY (matching) angle in the triangle to be equal. The only exception I can come up with, is when one of the triangles is "mirrored".

The SSA theorem (side side angle) does not ALWAYS prove congruency. The inclusion of Trigonometry may or may not prove this particular pair of triangles to be congruent. The fact that the long horizontal line is made up of two congruent (equal length) means bisection proof rules apply. The opposing angles of bisection are congruent and therefore you have the ingredients for an SSA proof BUT SSA is not a definitive theorem to prove congruency. As IOI said, the choice of lines that have been given as congruent along with the intentionally missing right angle or congruency markers for the included angle are provided here to create an SSA ambiguity.

Look they are very much congruent. My English 12 teacher once said something very good about this. There may be more than one right answer but there is usually a BETTER answer, and they will mark you for finding that better answer. Teachers put those triangles like that for a reason. You can argue all you like about it but if he puts down that they aren't he will get the question wrong I guarantee you.

Ok - after further reading, I do believe you are correct.

I found a link to an article earlier today that showed that SSA is sufficient proof - "as long as the angle is opposite the longer of the two equal sides" (?). After looking at this on paper, I can see why SSA is generally not sufficient (what from I can see there are two possible solutions for any triangle given SSA - basically an "inner & "outer" solution).

And I agree with both of you now - the dataset given is intentionally given to expose this ambiguity.

The triangles are NOT congruent!

(interesting question :>)

This discussion makes my eyes bleed and sort of makes me realiZe that people really don't know much math... like at all.

Let's call the sides marked with one line as A, and those marked with two lines as B. Now, ASSUMING the lines in the diagram are straight, the two angles at the center of the figure have to be identical, as they are cuased by the interseccion of two straight lines. Lets say these angles have a value of ALPHA. Now, the Sin Rule says that A/sin(angle opposite A) = B/(sin opposite B) in a triangle. Therefore, since we know the anlge opposite both A's are the same (ALPHA), we therefore conclude that the angles opposibe B are also the same, let's call them BETA.

So then we have two triangles with angles ALPHA and BETA, therefore, since the sum of angles in a triangle is always 180°, the remaining angles must also be the same, let's call them GAMMA.

Now we apply the Cosine Rule: C^2 = A^2 + B^2 - 2*A*B*cos(GAMMA), since A, B, and GAMMA are the same for both triangles, we than have that both have the same other side lenght, C.

In conclusion, we have two triangles with angles ALPHA, BETA, GAMMA and with sides A, B, C, so both triangles are congruent, have a nice day.