Similar polygons are divided into similar triangles, and into triangles equal in multitude and in the same ratio as the wholes, and the polygon has to the polygon a ratio duplicate of that which the corresponding side has to the corresponding side.

Let *ABCDE* and *FGHKL* be similar polygons, and let *AB* correspond to *FG.*

I say that the polygons *ABCDE* and *FGHKL* are divided into similar triangles, and into triangles equal in multitude and in the same ratio as the wholes, and the polygon *ABCDE* has to the polygon *FGHKL* a ratio duplicate of that which *AB* has to *FG.*

Join *BE, CE, GL,* and *HL.*

Now, since the polygon *ABCDE* is similar to the polygon *FGHKL,* therefore the angle *BAE* equals the angle *GFL,* and *AB* is to *AE* as *GF* is to *FL.*

Since then *ABE* and *FGL* are two triangles having one angle equal to one angle and the sides about the equal angles proportional, therefore the triangle *ABE* is equiangular with the triangle *FGL,* so that it is also similar, therefore the angle *ABE* equals the angle *FGL.*

But the whole angle *ABC* also equals the whole angle *FGH* because of the similarity of the polygons, therefore the remaining angle *EBC* equals the angle *LGH.*

And, since the triangles *ABE* and *FGL* are similar, *BE* is to *AB* as *GL* is to *GF.* Also, since the polygons are similar, *AB* is to *BC* as *FG* is to *GH.* Therefore, *ex aequali, BE* is to *BC* as *GL* is to *GH,* that is, the sides about the equal angles *EBC* and *LGH* are proportional. Therefore the triangle *EBC* is equiangular with the triangle *LGH,* so that the triangle *EBC* is also similar to the triangle *LGH.*

For the same reason the triangle *ECD* is also similar to the triangle *LHK.*

Therefore the similar polygons *ABCDE* and *FGHKL* have been divided into similar triangles, and into triangles equal in multitude.

I say that they are also in the same ratio as the wholes, that is, in such manner that the triangles are proportional, and *ABE, EBC,* and *ECD* are antecedents, while *FGL, LGH,* and *LHK* are their consequents, and that the polygon *ABCDE* has to the polygon *FGHKL* a ratio duplicate of that which the corresponding side has to the corresponding side, that is *AB* to *FG.*

Join *AC* and *FH.*

Then since the polygons are similar, the angle *ABC* equals the angle *FGH,* and *AB* is to *BC* as *FG* is to *GH,* the triangle *ABC* is equiangular with the triangle *FGH,* therefore the angle *BAC* equals the angle *GFH,* and the angle *BCA* to the angle *GHF.*

And, since the angle *BAM* equals the angle *GFN,* and the angle *ABM* also equals the angle *FGN,* therefore the remaining angle *AMB* also equals the remaining angle *FNG.* Therefore the triangle *ABM* is equiangular with the triangle *FGN.*

Similarly we can prove that the triangle *BMC* is also equiangular with the triangle *GNH.*

Therefore, proportionally, *AM* is to *MB* as *FN* is to *NG,* and *BM* is to *MC* as *GN* is to *NH.* So that, in addition, *ex aequali, AM* is to *MC* as *FN* is to *NH.*

But *AM* is to *MC* as the triangle *ABM* is to *MBC,* and as *AME* is to *EMC,* for they are to one another as their bases.

Therefore also one of the antecedents is to one of the consequents as are all the antecedents to all the consequents, therefore the triangle *AMB* is to *BMC* as *ABE* is to *CBE.*

But *AMB* is to *BMC* as *AM* is to *MC,* therefore *AM* is to *MC* as the triangle *ABE* is to the triangle *EBC.*

For the same reason also *FN* is to *NH* as the triangle *FGL* is to the triangle *GLH.*

And *AM* is to *MC* as *FN* is to *NH,* therefore the triangle *ABE* is to the triangle *BEC* as the triangle *FGL* is to the triangle *GLH,* and, alternately, the triangle *ABE* is to the triangle *FGL* as the triangle *BEC* is to the triangle *GLH.*

Similarly we can prove, if *BD* and *GK* are joined, that the triangle *BEC* is to the triangle *LGH* as the triangle *ECD* is to the triangle *LHK.*

And since the triangle *ABE* is to the triangle *FGL* as *EBC* is to *LGH,* and further as *ECD* is to *LHK,* therefore also one of the antecedents is to one of the consequents as the sum of the antecedents to the sum of the consequents. Therefore the triangle *ABE* is to the triangle *FGL* as the polygon *ABCDE* is to the polygon *FGHKL.*

But the triangle *ABE* has to the triangle *FGL* a ratio duplicate of that which the corresponding side *AB* has to the corresponding side *FG,* for similar triangles are in the duplicate ratio of the corresponding sides.

Therefore the polygon *ABCDE* also has to the polygon *FGHKL* a ratio duplicate of that which the corresponding side *AB* has to the corresponding side *FG.*

Therefore, *similar polygons are divided into similar triangles, and into triangles equal in multitude and in the same ratio as the wholes, and the polygon has to the polygon a ratio duplicate of that which the corresponding side has to the corresponding side.*

Q.E.D.

Similarly also it can be proved in the case of quadrilaterals that they are in the duplicate ratio of the corresponding sides. And it was also proved in the case of triangles, therefore also, generally, similar rectilinear figures are to one another in the duplicate ratio of the corresponding sides.