Pythagorean theorem from Heron's formula: Another proof

TL;DR

This paper presents an elementary geometric proof of the Pythagorean theorem derived from Heron's formula, accessible at a school-level understanding of geometry.

Contribution

It offers a novel, elementary proof of the Pythagorean theorem based on Heron's formula, simplifying the understanding of this fundamental relation.

Findings

Elementary proof of Pythagorean theorem derived from Heron's formula

Accessible geometric proof suitable for school-level education

Reinforces understanding of classical geometry concepts

Abstract

In this article using elementary school level Geometry we observe an alternative proof of Pythagorean Theorem from Heron's Formula.

Equations

$${\overline{AB}}^2+{\overline{AC}}^2={\overline{BC}}^2.$$

$$TheareaoftherectangleABDC$$

$$ab=|a\sqrt{r^2-\frac{a^2}{4}}|+|b\sqrt{r^2-\frac{b^2}{4}}|.$$

$$2ab=|a\sqrt{t+b^2}|+|b\sqrt{t+a^2}|,$$

$$2a^2{b}^2-(a^2+b^2)t=2ab|\sqrt{t^2+(a^2+b^2)t+a^2{b}^2}|$$

$$(a^2-b^2{)}^2{t}^2-8a^2{b}^2(a^2+b^2)t=0.$$

$$|a\sqrt{t+b^2}|+|b\sqrt{t+a^2}|>2ab,$$

Full text

††footnotetext: 2010 Mathematics Subject Classification: 51M04,51-03,01A20.

1. Pythagorean Theorem from Heron’s Formula -Another Proof

For a right angled triangle ABC (where $\angle BAC$ is the right angle), we have to show that

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For the proof, let O be the mid point $\overline{BC}$. Since $\angle BAC$ is the right angle, we can draw a circle with center O and radius $\overline{BO}$ such that the points $A,B,C$ lies on the circle. Now we join two points $A$ and $O$ and draw a line segment from the point B parallel to AC which meets the circle at D, (say). Again we join the points C,D and O,D.

Assume $\overline{AB}=a,~{}\overline{AC}=b,~{}\overline{BO}=r$.

Clearly from the figure,

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Now the area of the triangle ABO =$|\sqrt{s(s-a)(s-r)(s-r)}|=|\frac{a}{2}\sqrt{r^{2}-\frac{a^{2}}{4}}|,$

where $s=\frac{a+r+r}{2}$, and $|x|$ denotes the absolute value of $x$.

Similarly the area of the triangle AOC= $|\frac{b}{2}\sqrt{r^{2}-\frac{b^{2}}{4}}|.$

Thus from 1,

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By putting $t=4r^{2}-a^{2}-b^{2}$ in above equation we have

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i.e.,

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Again squaring both sides, we get

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Now we consider two cases :

Case-1 $a=b$

Then clearly $t=0$, i.e., $4r^{2}=a^{2}+b^{2}$.

Case-2 $a\not=b$

In this case we have $t\geq 0$.

Now when $t>0$, then

[TABLE]

which contradicts the equation (1.2).

So $t=0$, i.e., $4r^{2}=a^{2}+b^{2}$.

Hence $\overline{AB}^{2}+\overline{AC}^{2}=a^{2}+b^{2}=(2r)^{2}=\overline{BC}^{2}$. Hence the proof.