Pythagoras’ Theorem used to prove a triangle is right angled

Question

I was asked to prove that the vertices of a triangle, namely $A = 3 +i$, $B = 6$ and $C = 4 + 4i$ form a right angled triangle. I went down the path of showing that the squaring the two shorter sides and adding them, equals to the longest side, squared.

This got me thinking that how can we say that no other triangle will satisfy this equation since Pythagoras' theorem simply tells us that the The sum of the areas of the two squares on the legs (a and b) equals the area of the square on the hypotenuse (c).

Just because this equality stands for right angled triangles, how can we assume that it won't stand for others?

Is there a more rigorous method to show this?

Answer 1

You're right that the fact that the Pythagoras theorem is true, doesn't prove its converse must also be true. But, in this case it turns out that it can be easily proved that if Pythagoras's theorem is true then the converse is also true. The following simple proof is from the Wikipedia article:

Let ABC be a triangle with side lengths $a$, $b$, and $c$, with $a^2 + b^2 = c^2$. Construct a second triangle with sides of length $a$ and $b$ containing a right angle. By the Pythagorean theorem, it follows that the hypotenuse of this triangle has length $c = \sqrt{a^2 + b^2}$, the same as the hypotenuse of the first triangle. Since both triangles' sides are the same lengths $a$, $b$ and $c$, the triangles are congruent and must have the same angles. Therefore, the angle between the side of lengths $a$ and $b$ in the original triangle is a right angle.