# The Generalised Pythagoras Theorem – Another Proof

In this short note, we present a proof of the generalised Pythagoras theorem. We use the ‘ordinary’ Pythagoras theorem for the proof.

Theorem. In any triangle ABC, we have:

AC2 + BC2 > AB2 <--> C < 90 ,

AC2 + BC2 < AB2  <--> C > 90◦ .

Proof. On the coordinate plane, place the triangle ABC so that vertex C lies at the origin, side CB lies along the positive x-axis, and vertex A lies in the upper half plane (i.e., in the first or the second quadrants); see Figure 1. Let the coordinates of the three vertices be as follows: C = (0, 0); B = (r, 0), where r > 0; and A = (s, t), where t > 0. Then we have:                                BC2 = r 2, and AC2 = s2 + t 2, so

BC 2+ AC2 = r 2 + s 2 + t 2

Also:

AB2 = (r − s)2  + t2  = r2 + s 2 + t 2 − 2rs,

so

BC2 + AC2 − AB2 = 2rs.

Now it is clear from Figure 1 that:

C < 90◦ <--> s > 0,

C > 90◦ <--> s < 0.

Note that 2rs has the same sign as s (since r > 0). It follows that

BC2 + AC2 > AB2 <--> C < 90◦,

BC2 + AC2 < AB2 <--> C > 90◦.

We may thus state the “generalised Pythagoras theorem” as follows.

Theorem (Generalised Pythagoras theorem). In any triangle ABC, we have:

AC 2+ BC2 > AB2 <-->

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