Ratio of Areas of Similar Triangles

Theorem
That is, the ratio of the areas of the similar triangles is the square of the ratio of the corresponding sides.

Proof
Let $\triangle ABC$ and $\triangle DEF$ be similar, such that $\angle ABC = \angle DEF$ and $AB : BC = DE : EF$ such that $BC$ corresponds to $EF$.


 * Euclid-VI-19.png

Let $BG$ be constructed such that $EF : BG = BC : EF$, and join $AG$.

From Proportional Magnitudes are Proportional Alternately $AB : DE = BC : EF$.

So from Equality of Ratios is Transitive $AB : DE = EF : BG$.

So in $\triangle ABC$ and $\triangle DEF$ the sides about the equal angles are reciprocally proportional.

From Sides of Equiangular Triangles are Reciprocally Proportional‎, the area of $\triangle ABG$ equals the area of $\triangle DEF$.

Now we have that $BC : EF = EF : BG$.

So from $BC$ has to $BG$ a ratio duplicate to that which $CB$ has to $EF$.

But from Areas of Triangles and Parallelograms Proportional to Base, $CB : BG = \triangle ABC : \triangle ABG$.

So $\triangle ABC$ has to $\triangle ABG$ a ratio duplicate to that which $BC$ has to $EF$.

But $\triangle ABC = \triangle DEF$.

So $\triangle ABC$ has to $\triangle DEF$ a ratio duplicate to that which $BC$ has to $EF$.