# The two-circles method of proving the Pythagorean Theorem

The Pythagorean theorem, stating that the square of the hypotenuse is equal to the sum of the squares of the sides in a right triangle, is a fundamental concept in geometry and trigonometry.

There are many, many ways to prove the Pythagorean theorem. This page contains 84 different proofs, and The Pythagorean Proposition by Loomis contains another 367 proofs.

This is what I think to be an interesting proof, by the use of two circles. It is number 89 in Loomis’s book.

Starting with a right triangle, we draw two circles: Here, $\angle ACB$ is right, and A and B are the centers of circles $\Omega_1$ and $\Omega_2$ respectively.

Now we draw some additional lines, extending AB to F and G: In this diagram, we can prove that $\triangle ADC \sim \triangle ACG$:

1. $\angle DCG$ is right. This is an application of Thales’ theorem since DG is a diameter.
2. $\angle ACB$ is right. This is given.
3. $\angle ACD = \angle BCG$. Since $\angle DCG = \angle ACB$, $\angle DCG - \angle DCB = \angle ACB - \angle DCB$.
4. $\angle BCG = \angle BGC$. This is because BC and BG are both radii of $\Omega_2$ and $\triangle BCG$ is isosceles.
5. $\angle ACD = \angle BGC = \angle AGC$.
6. $\therefore \triangle ADC \sim \triangle ACG$. The two triangles have two shared angles: $\angle CAG$ and $\angle ACD = \angle AGC$.

In a similar way, we can prove that $\triangle BEC \sim \triangle BCF$.

The rest of the proof is algebraic rather than geometric. Let’s call the side AC to be b, BC=a, and AB=c.

From the similar triangles, we have the following ratios: $\frac{AG}{AC} = \frac{AC}{AD}$ (or, $b^2 = AG \cdot AD$) $\frac{BF}{BC} = \frac{BC}{EB}$ (or, $a^2 = BF \cdot EB$)

Adding the two equations, we get: $a^2 + b^2 = BF \cdot EB + AG \cdot AD$

The line BF can be split into AF and AB which is equal to c+b since AF = AC.

The line EB can be considered the difference between AB and AE, which is equal to c-b.

Similarly, AG = AB+BG = c+a, and AD = AB-DB = c-a. By substitution: $\begin{array}{rcl} a^2 + b^2 &=& (c+b) (c-b) + (c+a) (c-a) \\ &=& c^2 - b^2 + c^2 - a^2 \\ &=& 2c^2 - a^2 - b^2 \\ 2a^2 + 2b^2 &=& 2c^2 \\ a^2 + b^2 &=& c^2 \end{array}$

Q.E.D.

## 4 thoughts on “The two-circles method of proving the Pythagorean Theorem”

1. navya kapoor says:

please show some methods or some points by which ican prove pythagoras theorem in circles or e – mail it on my e – mail address navukappor@gmail.com

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1. himika singh says:

yes i also want these points to make my project report

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2. Anonymous says:

thanks a lot it was very useful n easy to understand

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