[Math] geometry triangles side-side-side | prove the teacher she is wrong

geometrytriangles

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?

Best Answer

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.

triangles

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.

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