For let AC, FH be joined.
Then since, because of the similarity of the polygons, the angle ABC is equal to the angle FGH,
and, as AB is to BC, so is FG to GH,
the triangle ABC is equiangular with the triangle FGH; [VI. 6]
therefore the angle BAC is equal to the angle GFH, and the angle BCA to the angle GHF.
And, since the angle BAM is equal to the angle GFN,
and the angle ABM is also equal to the angle FGN,
therefore the remaining angle AMB is also equal to the remaining angle FNG; [I. 32]
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, as AM is to MB, so is FN to NG,
and, as BM is to MC, so is GN to NH; so that,
in addition, ex aequali, as AM is to MC, so is FN to NH.
But, as AM is to MC, so is the triangle ABM to MBC, and AME to EMC;
for they are to one another as their bases. [VI. 1]
Therefore also, as one of the antecedents is to one of the consequents, so are all the antecedents to all the consequents; [V. 12]
therefore, as the triangle AMB is to BMC, so is ABE to CBE.
But, as AMB is to BMC, so is AM to MC;
therefore also, as AM is to MC, so is the triangle ABE to the triangle EBC.
For the same reason also, as FN is to NH, so is the triangle FGL to the triangle GLH.
And, as AM is to MC, so is FN to NH;
therefore also, as the triangle ABE is to the triangle BEC, so is the triangle FGL to the triangle GLH;
and, alternately, as the triangle ABE is to the triangle FGL, so is the triangle BEC to the triangle GLH.
Similarly we can prove, if BD, GK be joined,
that, as the triangle BEC is to the triangle LGH, so also is the triangle ECD to the triangle LHK.
And since, as the triangle ABE is to the triangle FGL, so is EBC to LGH,
and further ECD to LHK,
therefore also, as one of the antecedents is to one of the consequents so are all the antecedents to all the consequents; [V. 12]
therefore, as the triangle ABE is to the triangle FGL, so is the polygon ABCDE 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. [VI. 19]
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.
Complete text
Book I