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Believe it or not, there are over 200 different proofs for the Pythagorean theorem. It is one of the most studied aspects of mathematics!

Here is the proof I think is easiest—it is a direct proof. Enjoy!

Example 1

From the drawing, you can see that there are two ways to describe the area of the large square. Either as:

length times width for the large square

(length = $a+b$ and width = $a+b$),

or as

the sum of the areas of the small square and the four right-angled triangles.

Since these two approaches provide the same answer— namely—the area of the large square, you can make an equation. One side of the equation is “length times the width of the large square”, and the other side of the equal sign is “the sum of the areas of the small square and the four right triangles”. Therefore,

$${A}_{\text{largesquare}}={A}_{\text{smallsquare}}+4\cdot {A}_{\text{triangle}}$$ |

To be able to insert expressions into the equation, you need to describe these areas using the figure above: $$\begin{array}{llll}\hfill {A}_{\text{largesquare}}& =\text{length}\cdot \text{width}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\phantom{\rule{-0.17em}{0ex}}\left(a+b\right)\phantom{\rule{-0.17em}{0ex}}\left(a+b\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={a}^{2}+2ab+{b}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {A}_{\text{smallsquare}}& =\text{length}\cdot \text{width}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =c\cdot c\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={c}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {A}_{\text{triangle}}& =\frac{g\cdot h}{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{a\cdot b}{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$

You now insert the expressions into the equation above, and get: $$\begin{array}{llll}\hfill {a}^{2}+2ab+{b}^{2}& ={c}^{2}+4\cdot \frac{a\cdot b}{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {a}^{2}+2ab+{b}^{2}& ={c}^{2}+2ab\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {a}^{2}+{b}^{2}& ={c}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$

And you have proven the Pythagorean theorem just like that!

Q.E.D