# Pythagoras' Theorem used to prove a triangle is right angled

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?

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.

You are asking for the converse of the usual, direct, formulation of the Pythagorus Theorem. You are right that the direct theorem does not imply its converse.

The fact is that the converse is true too. In my daughter's maths class (in France), they teach explicitly the theorem and its converse.

The question about the converse of the Pythagorean theorem has been adequately addressed in the other answers and comments. I am writing this answer to give you a simpler and shorter solution to the original question.

You are dealing with the complex plane. To show that the points make a right angled triangle, you need to show that two of the sides are perpendicular. In the complex plane, to show that the vector representing a complex number $$z$$ is perpendicular to the vector representing another complex number $$w$$, you need to show that $$z = kiw$$, where $$k$$ is some real number. In particular, when $$k = 1$$, you have $$|z| = |w|$$ (equal lengths) and that $$z$$ is constructed from a counterclockwise $$90$$ degree rotation from $$w$$, and when $$k=-1$$, it's a clockwise $$90$$ degree rotation instead. Where $$|k| \neq 1$$, you have a length scaling factor.

That said, find the vectors $$\vec {AB}, \vec {BC}$$ and $$\vec{AC}$$ in complex form:

$$\vec {AB} = \vec{OB} - \vec{OA} = 6 - 3- i = 3-i$$

Similarly, $$\vec {BC} = -2+4i$$ and $$\vec{AC} = 1+3i$$

Almost immediately, you should be able to "see" that $$\vec{AC} = 1+3i = i(3-i) = i\vec{AB}$$.

Hence you can conclude that $$A,B,C$$ form a right triangle in the complex plane with $$AC$$ and $$AB$$ being perpendicular (the right angle being at $$A$$), of equal length (isosceles right triangle), and with segment $$AC$$ constructible with a $$90$$ degree counterclockwise rotation of segment $$AB$$.

The law of cosines tells us that for a triangle with sides $$a$$, $$b$$ and $$b$$ and angles $$\alpha$$, $$\beta$$ and $$\delta$$, we have that

$$c^2 = a^2 + b^2 - 2ab \cos \gamma$$

So if we know that $$c^2 = a^2 + b^2$$, we know that $$\cos \gamma = 0$$. This can only be the case if $$\gamma = \frac{\pi}{2}$$, since we know that $$0 < \alpha,\beta,\gamma < \pi$$.

In the complex plane, the points are : $$A(3,1) , B(6,0)$$ and $$C(4,4)$$. Let $$m_l$$ denotes the slope of any line $$l$$ .

Observe that : $$m_{AB} . m_{AC}=-1\implies \angle BAC=\frac{\pi}{2}$$

A quick solution is to prove that if $$AB^2+AC^2 = BC^2$$ then $$\Delta ABC$$ is right at $$A$$ is to construct the altitude $$AH$$. Then $$AB^2+AC^2 = 2HA^2+HB^2+HC^2$$ using Pythagorean theorem. Then based on the fact that $$AB^2+AC^2 = BC^2 = (HB+HC)^2 = HB^2+2HB\cdot HC+HC^2$$ we arrive at $$HB\cdot HC = HA^2$$, or $$HB/HA = HA/HC$$. This means the two right triangles $$\Delta BHA$$ and $$\Delta AHC$$ are similar. Consequently $$\angle BAC = \angle BAH + \angle HAC = \angle ACH + \angle BCH = 90^\circ$$

the slope of AC = (4-1)/(4-3) = 3 the slope of AB = (1-0)/(3-(-6)) = -1/3 since the 2 slopes are negative reciprocals of each other, then AC is perpendicular to AB, thus right angle