Theorem of Pythagoras demonstration and its geometric illustration (using the Thales theorem)

In this publication will be made a formal proof of the Pythagorean theorem, using the theorem of Thales, together with an geometric illustration of the results that are being achieved.

Consider the triangle $ \left [ABC \right] $ rectangle in $ C $ and $ \left [CD \right] $ its height relative to the side $ \left [AB \right] $. Let $ a = \overline{BC} $, $ b = \overline{AC} $, $ c = \overline {AB} $, $ x = \overline {AD} $ and $ y = \overline {DB} $ .

Step 1: Prove that the triangle $ \left [ABC \right] $ is similar to the triangle $ \left [ADC \right] $ and the triangle $ \left [ABC \right] $ is similar to the triangle $ \left [BDC \right] $.

• As the triangle $\left [ABC \right] $ and the triangle $ \left [ADC \right] $ are both rectangles and have the included angle of the vertex $A$ in common, by the AA criterion of similar triangles, the two triangles are similar.
• Similarly, as the triangle $\left [ABC \right] $ and the triangle $ \left [BDC \right] $ are both rectangles and have the included angle of the vertex $ B $ in common, by the AA criterion of similarity of triangles, two triangles are similar.

Step 2: Prove that $ \frac{\overline {AC}}{\overline {AB}} = \frac{\overline{AD}}{\overline {AC}}$ and $ \frac{\overline {BC} }{\overline {AB}} = \frac {\overline{DB}}{\overline{BC}} $. 

As by the first step we conclude that the triangle $\left [ABC \right] $ is similar to the triangle $\left [ADC \right] $ and 

• $\left [AC \right] $ corresponds to $\left[AB \right] $;
• $\left [AD \right] $ corresponds to $\left[AC \right] $. 

Then $$ \frac{\overline{AC}}{\overline{AB}} = \frac{\overline{AD}}{\overline{AC}} $$. 

Similarly, as at step 1 we conclude that the triangle $\left [ABC \right] $ is similar to the triangle $ \left [BDC \right]$ and 

• $\left[BC \right] $ corresponds to $ \left[AB \right] $;
• $\left[DB \right] $ corresponds to $ \left[BC \right] $.

Then $$\frac{\overline{BC}}{\overline{AB}} = \frac{\overline{DB}}{\overline{BC}}$$.

Step 3: Given the figure below: conclude that $b^{2} = x \times c$, that is, the area of the square $ \left[ACKL \right]$ is equal to the rectangle area $\left [ADOP \right]$; And conclude that $a^{2} = y \times c$, that is, the area of the square $\left [BCNM \right]$ is equal to the rectangle area $\left[DBOQ\right]$.

As in step 2 we conclude that $$\frac{\ overline{AC}}{\overline{AB}} = \frac{\overline{AD}}{\overline{AC}}$$ then, 

$$\frac{b}{c} = \frac{x}{b}. $$ Therefore, $ b ^ {2} = x \times c $.

Similarly as in step 2, we conclude that, $$\frac{\overline{BC}}{\overline{AB}} = \frac{\overline {DB}}{\overline {BC}}$$ then $$\frac{a}{c} = \frac{y}{a}. $$ 

So $a^{2} = y \times c $.

Step 4: Prove that $ a^{2} + b^{2} = c^{2} $. That is, the area of squares that are on the sides is equal to the square on the hypotenuse area.

After performing these four steps, we conclude intuitively that the sum of the area of the square $\left[ACKL\right] $ with the area of the square $\left[BCNM\right]$ is equal to the area of the square $\left[ACKL\right] $. 

However, being more rigorous, at step 3, because $b^{2} = x \times c $ and $ a^{2} = y \times c $, then $$ a^{2} + b^{2} = y \times c + x \times c = (y + x) \times c = c \times c = c^{2} $$ because $ x + y = c $. Therefore, the $ a^{2} + b^{2} = c^{2}$.

The following application Geogebra illustrates perfectly all this demonstration.