Ratio of Areas of Similar Triangles

Theorem

 * Similar triangles are to one another in the duplicate ratio of the corresponding sides.

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 $EF$.

Porism

 * From this it is manifest that, if three straight lines be proportional then, as the first is to the third, so is the figure described on the first to that which is similar and similarly described in the second.