What’s the way to prove that its two corresponding sides are equal that <i>isn’t</i> by contradiction?
Are you sure you have to prove the sides equal and not the base angles? Because the very definition of an isosceles triangle is that it has at least two equal sides. But anyway, assuming you wanted to prove it ANYWAY, you could prove that congruence of angles implies congruence of sides.
If you have a triangle XYZ, where angle Y = angle Z, and thus angle Z = angle Y. Thus side YZ=ZY. Then, using angle-side-angle to prove two triangles congruent, triangle ZXY = triangle YXZ. Thus XZ=XY.
I wasn’t sure if I was allowed to use ASA, since that seems to be on a higher level than this…
I don’t know of any proofs that don’t use ASA. You’d use SAS to prove the converse of that theorem, by the way, that congruent sides imply congruent angles.
What axioms, theorems, etc., are you allowed to use that you don’t need to prove yourself? Trivial math problems are really annoying.
Well, in Euclidian geometry there are the four postulates of Euclid (fifth if you really want to get into it), so…
a) a straight line segment can be drawn between any two points
b) a straight line segment can be extended infinitely in a straight line
c) a circle of 360 degrees may be drawn from a line segment given that one end is the center
d) all right angles are equal (but some right angles are more equal than other, my comrades)
e) if a straight line intersecting two other lines such that the inside angles do not form perfect right angles, then the aforementioned two other lines must cross at some point extended beyond the side where the sum of the two inside angles are <i>less</i> than 180 degrees. i.e. two lines which are not parallel on a 2-D plane must meet one another on any given plane.
If you want to delve into non-Euclidian geometry be my guest 