# Pythagorean Theorem by Analytic Geometry

### Nuno Luzia

Universidade Federal do Rio de Janeiro,

Instituto de Matemática

Rio de Janeiro 21941-909, Brazil

The theorem is ilustrated in figure below.

$a^2+b^2=c^2$

The figure below is obtained by putting another right triangle (marked with dotted line) on top of the original one, so as to form a rectangle with sides with lengths $a$ and $b$, and then reflecting it through the common hypotenuse.

We get an isosceles triangle with base length $c$ and height $h$, and two smaller right rectangles with sides of lengths $l$, $b$ and $a-l,$ for some $l.$

The area of the isosceles triangle is equal to the area of the original triangle minus the area of one smaller right rectangle. So

(1)

$ch=b(a-l).$

(This equation can also be obtained by using a similarity between two right rectangles.)

We will determine $h$ and $l$ by using cartesian coordinates and basic vector calculus. We put the origin $O$ on the left corner of the original triangle, with the usual $x$ and $y$ axis for the cartesian coordinates. In this way, the vector $\vec{OP}$ along the hypotenuse is represented by the coordinates $(a,b)$.

To determine $h$ and $l$ we should find an orthogonal vector to $(a,b)$. Geometrically, changing the roles of $x$ and $y$ axis corresponds to a rotation of $90^\circ$ followed by a reflection through the horizontal axis, so the vector $(b,-a)$ is orthogonal to $(a,b)$.

Then the straight line segment starting at the midpoint of $OP$ and orthogonal to it is represented by

$(a/2, b/2)+t (b,-a),\quad t\ge 0.$

The $y$ coordinate of expression above is zero when $\displaystyle t=\frac{b}{2a}$, and then the $x$ coordinate is $\displaystyle \frac{a}{2}+\frac{b^2}{2a}$. So

(2)

$\displaystyle a-l=\frac{a}{2}+\frac{b^2}{2a}.$

Now $h$ is the length of $\displaystyle \frac{b}{2a} (b,-a)$ which is the length of $\displaystyle \frac{b}{2a} OP$. So

(3)

$\displaystyle h=\frac{b}{2a} c.$

(This equation can also be obtained by using a similarity between two right rectangles.)

Putting (2) and (3) into (1) we obtain the Pythagorean relation.

(**Note** that the validity of using calculus, let alone analytic geometry, to prove the Pythagorean theorem has been argued elsewhere.)

|Pythagorean Theorem| |Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny70786464