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

Let *ABC* and *DEF* be similar triangles having the angle at *B* equal to the angle at *E,* and such that *AB* is to *BC* as *DE* is to *EF,* so that *BC* corresponds to *EF.*

I say that the triangle *ABC* has to the triangle *DEF* a ratio duplicate of that which *BC* has to *EF.*

Take a third proportional *BG* to *BC* and *EF* so that *BC* is to *EF* as *EF* is to *BG,* and join *AG.*

Since *AB* is to *BC* as *DE* is to *EF,* therefore, alternately, *AB* is to *DE* as *BC* is to *EF.*

But *BC* is to *EF* as *EF* is to *BG,* therefore also *AB* is to *DE* as *EF* is to *BG.*

Therefore in the triangles *ABG* and *DEF* the sides about the equal angles are reciprocally proportional.

But those triangles which have one angle equal to one angle, and in which the sides about the equal angles are reciprocally proportional, are equal. Therefore the triangle *ABG* equals the triangle *DEF.*

Now since *BC* is to *EF* as *EF* is to *BG,* and, if three straight lines are proportional, the first has to the third a ratio duplicate of that which it has to the second, therefore *BC* has to *BG* a ratio duplicate of that which *BC* has to *EF.*

But *BC* is to *BG* as the triangle *ABC* is to the triangle *ABG,* therefore the triangle *ABC* also has to the triangle *ABG* a ratio duplicate of that which *BC* has to *EF.*

But the triangle *ABG* equals the triangle *DEF,* therefore the triangle *ABC* also has to the triangle *DEF* a ratio duplicate of that which *BC* has to *EF.*

Therefore, *similar triangles are to one another in the duplicate ratio of the corresponding sides.*

Q.E.D.

From this it is clear that if three straight lines are proportional, then the first is to the third as the figure described on the first is to that which is similar and similarly described on the second.