Area Proofs of Pythagoras

Introduction

Well, well, well. One of Pythagoras' famous theorems. It states that in a right triangle, the two lesser sides, $a$ and $b$, and the hypotenuse (the longest side) $c$ have the relationship $a^2+b^2=c^2$. So weird, how something so simple can open so many doors to geometry, like trigonometry and others.

Proof using areas

Here, we have square $DEFG$ of side length $a+b$ with a smaller square $HIJK$ of side length $c$.

We can write the area of the larger square $DEFG$ in two ways as you answered: $(a+b)^2$ and $4\cdot\dfrac{ab}{2} + c^2$. The former is the simple square area (side length squared) and the latter is the inner square area $HIJK$ plus four right triangles ($IFJ$, $JGK$, $KDH$, and $HEI$ are all congruent). Equating these two expressions, we find $$(a+b)^2 = 2ab+c^2.$$ Expanding the left side, this results in $$a^2+2ab+b^2 = 2ab+c^2,$$ which is $$a^2+b^2 = c^2!!!!!!!!!$$ Yay we proved Pythagorean theorem.