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.



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

This new figure is shown below.



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

QED

Proof 2


Let be $$\triangle ABC$$ 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.