First time I'm here, I'M REALLY frustrated by now. So I'll just give u the question first.

      / | \
     /  |  \       
    /  ---  \
   /    |    \
     ||   ||

  --- = congruent dash
  || = congruent dash
  ) or ( = congruent angles (70 degrees)

(Sorry for this triangle, I tried uploading a pic but I am new so I can't..)

So my question is, since we need to check if this is s-s-s, and we know the middle line is congruence in both triangle, and we were given that the base was congruent. We were also given two congruent angles (70 degrees)

So here is the big question.
My teacher says that the angles Are just a distraction.
What I'm saying is that since we know that both bases and one line is congruent, and the angle is the same in both triangles, will the last side also be congruent?

Am I right? Or is my teacher?

  • $\begingroup$ First, welcome on Math.SE! What does it mean for a triangle to be s-s-s? And the difference between your opinion and your teacher's is that you think that the congruence of the angles is important for the problem while your teacher says it doesn't matter? Is the vertical line assumed to be perpendicular to the base of the whole triangle? $\endgroup$ Commented Oct 10, 2012 at 14:59
  • 1
    $\begingroup$ First of all thanks for the welcoming. Now sss means sidecside side congruence and her opinion is that there is not enough info to decide if the last line is congruence or not. And the middle line bisect the base $\endgroup$
    – Baruch
    Commented Oct 10, 2012 at 15:01
  • $\begingroup$ It seems that s-s-s means congruence by three sides. The question is weather the base line assumed to be straight. $\endgroup$ Commented Oct 10, 2012 at 15:02
  • 1
    $\begingroup$ i50.tinypic.com/vzc1lx.jpg This is the picture $\endgroup$
    – Baruch
    Commented Oct 10, 2012 at 15:06
  • 1
    $\begingroup$ No problem, thank you guys for trying to help, I appreciate it. $\endgroup$
    – Baruch
    Commented Oct 10, 2012 at 15:08

3 Answers 3


I think there are a few points to be made here. First of all, let me clarify that you are correct, but perhaps not directly.

You are given a triangle with two base angles identical. That means the triangle is isosceles. You are also given a median of the triangle, which means that the median partitions the isosceles triangle into two right triangles. The subtlety here is that the above facts are not freely given. They require proof. If you have not shown that your median intersects your base at a right angle, then you cannot just assume the triangles are congruent.


Here is a picture with two triangles, $\triangle ABD$ and $\triangle ABE$. They share two sides of the same length, namely $\overline{AB} = \overline{AB}$ and $\overline{EB} = \overline{DB}$. They also share angle $\angle DAB = \angle EAB$. The triangles are not congruent.

There is sometimes a danger in geometry where too much is assumed from a given picture. Certainly pictures are indispensable in geometry, but always keep in mind that they are merely guides. This becomes especially true in higher geometry, where it's no longer possible (or feasible) to accurately illustrate all aspects of the problem (for example, inverse geometry or hyperbolic geometry becomes a little more difficult to picture).

When you solved your problem, you implicitly made use of side-angle-side congruence (note the angle has to be in between the two sides you use) using the base, the right angle and the shared median. That is perhaps the more natural congruence to use, but if you have not shown that the two triangles are indeed both right triangles, then you have skipped steps.

  • $\begingroup$ thanks for your time and effort. I still have one more question, since the bases are congruent, and the Middle line is congruent, it means the triangle has the same height right? So if that's is right, wouldn't the Angle on the sides be the one deciding the length of the the line on the sides? Like what i mean is kind of hard to explain. The triangle itself is given that it has two congruent lines and one congruent Angle which is on the external sides of the triangle so my hypothesis is that this angle will decide the size of these lines... sorry if i confused you guys. $\endgroup$
    – Baruch
    Commented Oct 10, 2012 at 16:27
  • $\begingroup$ I choose this answer since it helped me the most and had a lot of information. Thank you everybody else that participated and tried to help. $\endgroup$
    – Baruch
    Commented Oct 10, 2012 at 17:08
  • $\begingroup$ @Baruch The thing is, you can't simply decide that they have the same height. The example I gave has two triangles with two sides and an angle congruent, but they do not have the same height. I know this is a bit difficult to grasp, partly because your intuition tells you that they are the same, and even more so because your intuition is right in this case. But we are trying to teach you to stray away from intuition a bit and to use formal justifications, because there will be cases (many actually) where intuition fails you. $\endgroup$
    – EuYu
    Commented Oct 10, 2012 at 17:14
  • $\begingroup$ Another thing I might comment on. I disagree a bit with your teacher's comment that the angles are "distractions". I do not want you to think that angles are useless in geometry; they are central. I personally don't really believe in the notion of a red-herring problem. In "the real world", you never have just the right amount of facts to solve a problem. The more information you have the better. $\endgroup$
    – EuYu
    Commented Oct 10, 2012 at 17:22
  • $\begingroup$ Thanks for everything, now I understand much more about it. I had learned a lot from u guys. Thanks again $\endgroup$
    – Baruch
    Commented Oct 10, 2012 at 18:27

If the image represents a triangle, then you are correct. If the image represents two triangles sharing a side, then your teacher is correct.

Symbolically, you have triangles $\Delta ABC$ and $\Delta DBC$, where $A$ is the left vertex, $D$ is the right vertex, $B$ is the top vertex, and $C$ is the center vertex. You have the shared side $BC$, clearly, and are given that $\angle A \cong \angle D$. What isn't entirely clear is if $A$, $C$, and $D$ are collinear. If they are, then $\angle BCA$ and $\angle BCD$ are complementary. This additional fact is enough to prove congruence. If they aren't (which doesn't conflict with what you've said you're given, although the image in that case is a bit misleading), then the two triangles may not be congruent.


Not necessarily. See this page for some reference on what needs to be present for congruence, and also check out this section specifically, regarding the problem you have and the common slip up of wanting to jump to congruence with two sides and an angle (SSA) defined and why that isn't possible.

  a   /|\  b
     / | \
    / -d- \
   /   |   \
   c1     c2

The problem is that without being able to prove that line d either is perpendicular to line c OR bisects angle C, you don't have a second angle and no congruence proof can work. If you know either of these things, you now have enough information to capture that angle (either C/2 or the 90 degree with d and c) and pair it with the 70 degree A and B and then use almost any of the proofs with the amount of information you have and then you would be right. Unfortunately, without that, your teacher is right.

  • $\begingroup$ Since the given angles are 70°, necessarily $d$ is longer than $c_{1/2}$ and thus the halves are congruent. We see that it is important that the values of the angles are given (or at least the fact that they are $>45°$). Thus if the teacher says, the angles are not important, he/she is wrong. $\endgroup$ Commented Oct 15, 2012 at 22:16
  • $\begingroup$ @HagenvonEitzen Absolutely. Without that information, the case for congruence is severely crippled if not eliminated. Granted, I'm still blurry on the details of what is where and the givens from the "image" provided. $\endgroup$
    – jimcavoli
    Commented Oct 16, 2012 at 13:19

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