Pythagoras's Theorem

Theorem
Given any right triangle $$\triangle ABC$$ with $$c$$ as the hypotenuse, we have $$ a^2+b^2=c^2$$.

Proof 1
So, consider the triangle shown below.


 * Triangle.jpg

So, we can extend this triangle into a square by transforming it using isometries, specifically rotations and translations.

This new figure is shown below.


 * Square.jpg

So, this figure is clearly a square, since all the angles are right angles, and the lines connecting the corners are easily seen to be straight.

Now, let's calculate the area of this figure.

On the one hand, we have the area of the square as $$(a+b)^2=a^2+2ab+b^2$$.

On the other hand, we can add up the area of the component parts of the square, specifically, we can add up the four triangles and the inner square.

Thus we have the area of the square to be $$4\left({\frac{1}{2}ab}\right) + c^2=2ab+c^2$$.

Now these two expressions have to be equal, since they both represent the area of the square.

Thus, $$a^2+2ab+b^2=2ab+c^2 \iff a^2+b^2=c^2$$.

Proof 2

 * [[File:Phytagoras.PNG]]

Let $$\triangle ABC$$ be a right triangle and $$h_c$$ the altitude from $$c$$.

We have


 * $$\angle CAB \cong \angle DCB$$


 * $$\angle ABC \cong \angle ACD$$

Then we have


 * $$\triangle ADC \sim \triangle ACB \sim \triangle CDB$$

Use the fact that if $$\triangle XYZ \sim \triangle X'Y'Z'$$ then $$\frac{(XYZ)}{(X'Y'Z')}=\frac{XY^2}{X'Y'^2}=\frac{h_z^2}{h_{z'}^2}=\frac{t_z^2}{t_{z'}^2}=...$$ where $$(XYZ)$$ represents the area of $$\triangle XYZ$$.

This gives us $$\frac{(ADC)}{(ACB)} =\frac{AC^2}{AB^2}$$ and $$\frac{(CDB)}{(ACB)} = \frac{BC^2}{AB^2}$$.

Taking the sum of these two equalities we obtain $$\frac{(ADC)}{(ACB)}+ \frac{(CDB)}{(ACB)}=\frac{BC^2}{AB^2}+\frac{AC^2}{AB^2}$$.

Thus, $$\frac{\overbrace{(ADC)+(CDB)}^{(ACB)}}{(ACB)}=\frac{BC^2+AC^2}{AB^2}$$.

This gives us $$\therefore AB^2=BC^2+AC^2$$ as desired.

Algebraic Proof
We start with the algebraic definitions for sine and cosine:


 * $$\sin x = \sum_{n=0}^\infty \left({-1}\right)^n \frac {x^{2n+1}}{\left({2n+1}\right)!} = x - \frac {x^3} {3!} + \frac {x^5} {5!} - \cdots$$;


 * $$\cos x = \sum_{n=0}^\infty \left({-1}\right)^n \frac {x^{2n}}{\left({2n}\right)!} = 1 - \frac {x^2} {2!} + \frac {x^4} {4!} - \cdots$$.

From these, we derive the proof that $$\cos^2 x + \sin^2 x = 1$$.

Then from the Equivalence of Definitions for Sine and Cosine, we can use the geometric interpretation of sine and cosine:


 * SineCosine.png


 * $$\sin \theta = \frac {\text{Opposite}} {\text{Hypotenuse}}$$;
 * $$\cos \theta = =\frac {\text{Adjacent}} {\text{Hypotenuse}}$$.

Let $$\text{Adjacent} = a, \text{Opposite} = b, \text{Hypotenuse} = c$$, as in the diagram at the top of the page.

Thus:

$$ $$ $$

The Classic Proof

 * Euclid-I-47.png

Let $$ABC$$ be a right triangle whose angle $$BAC$$ is a right angle.

Construct squares $$BDEC$$ on $$BC$$, $$ABFG$$ on $$AB$$ and $$ACKH$$ on $$AC$$.

Construct $AL$ parallel to $$BD$$ (or $$CE$$).

Since $$\angle BAC$$ and $$\angle BAG$$ are both right angles, from Two Angles making Two Right Angles make a Straight Line it follows that $$CA$$ is in a straight line with $$AG$$.

For the same reason $$BA$$ is in a straight line with $$AH$$.

We have that $$\angle DBC = \angle FBA$$, because both are right angles.

We add $$\angle ABC$$ to each one to make $$\angle FBC$$ and $$\angle DBA$$.

By common notion 2, $$\angle FBC = \angle DBA$$.

By Triangle Side-Angle-Side Equality, $$\triangle ABD = \triangle FBC$$.

We have that the parallelogram $$BDLM$$ is on the same base $$BD$$ and between the same parallels $$BD$$ and $$AL$$ as the triangle $$\triangle ABD$$.

So, by Parallelogram on Same Base as Triangle has Twice its Area, the parallelogram $$BDLM$$ is twice the area of $$\triangle ABD$$.

Similarly, we have that the parallelogram $$ABFG$$ (which happens also to be a square) is on the same base $$FB$$ and between the same parallels $$FB$$ and $$GC$$ as the triangle $$\triangle FBC$$.

So, by Parallelogram on Same Base as Triangle has Twice its Area, the parallelogram $$ABFG$$ is twice the area of $$\triangle FBC$$.

So $$BDLM = 2 \triangle ABD = 2 \triangle FBC = ABFG$$.

By the same construction, we have that $$CELM = 2 \triangle ACE = 2 \triangle KBC = ACKH$$.

But $$BDLM + CELM$$ is the whole of the square $$BDEC$$.

Therefore the area of the square $$BDEC$$ is equal to the combined area of the squares $$ABFG$$ and $$ACKH$$.