Real Multiplication Distributes over Addition/Geometric Proof

Proof

 * Euclid-II-1.png

Let $A$ and $BC$ be two straight lines.

Let $BC$ be cut at random at points $D$ and $E$.

Then the rectangle contained by $A$ and $BC$ is equal to the sum of the rectangles contained by $A$ and $BD$, by $A$ and $DE$, and by $A$ and $EC$, as follows:

Construct $BF$ perpendicular to $BC$.

Construct $BG$ on $BF$ equal to $A$.

Construct $GH$ through $G$ parallel to $BC$.

Construct $DK, EL, CH$ through $D, E, C$ parallel to $BG$.

Then we have that:
 * $\Box BCHG = \Box BDKG + \Box DELK + \Box ECHL$

Now $\Box BCHG$ is the rectangle contained by $A$ and $BC$, because it is contained by $BC$ and $BG$, and $BG = A$.

Similarly, from Opposite Sides and Angles of Parallelogram are Equal:
 * $\Box BDKG$ is the rectangle contained by $A$ and $BD$, because it is contained by $BD$ and $BG = A$
 * $\Box DEKL$ is the rectangle contained by $A$ and $DE$, because it is contained by $DE$ and $DK$, and $DK = A$
 * $\Box ECHL$ is the rectangle contained by $A$ and $EC$, because it is contained by $EC$ and $EL$, and $EL = A$.

Hence the result.