Difference of Two Squares/Geometric Proof 1

Theorem

 * $\forall x, y \in \R: x^2 - y^2 = \paren {x + y} \paren {x - y}$

Proof

 * Euclid-II-5.png

Let $AB$ be cut into equal segments at $C$ and unequal segments at $D$.

Then the rectangle contained by $AD$ and $DB$ together with the square on $CD$ equals the square on $BC$.

(That is, let $x = AC, y = CD$. Then $\paren {x + y} \paren {x - y} + y^2 = x^2$.)

This is proved as follows.

Construct the square $CBFE$ on $CB$, and join $BE$.

Construct $DG$ parallel to $CE$ through $G$, and let $DG$ cross $BE$ at $H$.

Construct $KM$ parallel to $AB$ through $H$.

Construct $AK$ parallel to $BF$ through $A$.

From Complements of Parallelograms are Equal:
 * $\Box CDHL = \Box FGHM$.

Add the square $DBMH$ to each.

Then $\Box CBML = \Box DBFG$.

But as $AC = CB$, from Parallelograms with Equal Base and Same Height have Equal Area we have that:
 * $\Box ACLK = \Box CBML$

Add $\Box CDHL$ to each.

Then $\Box ADHK$ is equal in area to the gnomon $CBFGHL$.

But $\Box ADHK$ is the rectangle contained by $AD$ and $DB$, because $DB = DH$.

So the gnomon $CBFGHL$ is equal in area to the rectangle contained by $AD$ and $DB$.

Now $\Box LHGE$ is equal to the square on $CD$.

Add $\Box LHGE$ to each of the gnomon $CBFGHL$ and $\Box ADHK$.

Then the gnomon $CBFGHL$ together with $\Box LHGE$ equals the rectangle contained by $AD$ and $DB$ and the square on $CD$.

But the gnomon $CBFGHL$ together with $\Box LHGE$ is the square $CBFE$.

Hence the result.