Euclid, Elements
Thomas L. Heath, Sir Thomas Little Heath, Ed.

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But, as BC is to EF, so is EF to BG; therefore also, as AB is to DE, so is EF to BG. [V. 11]

Therefore in the triangles ABG, 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; [VI. 15]
therefore the triangle ABG is equal to the triangle DEF.

Now since, as BC is to EF, so is EF to BG, and, if three straight lines be proportional, the first has to the third a ratio duplicate of that which it has to the second, [V. Def. 9] therefore BC has to BG a ratio duplicate of that which CB
has to EF.

But, as CB is to BG, so is the triangle ABC to the triangle ABG; [VI. 1] therefore the triangle ABC also has to the triangle ABG a ratio duplicate of that which BC has to EF.

But the triangle ABG is equal to the triangle DEF; therefore the triangle ABC also has to the triangle DEF a ratio duplicate of that which BC has to EF.

Therefore etc.

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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 on the second. Q. E. D.

¹

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PROPOSITION 20.

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, FGHKL be similar polygons, and let AB correspond to FG; I say that the polygons ABCDE, 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.

Let BE, EC, GL, LH be joined.

Now, since the polygon ABCDE is similar to the polygon
FGHKL, the angle BAE is equal to the angle GFL;

and, as BA is to AE, so is GF to FL. [VI. Def. 1]

Since then ABE, 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; [VI. 6]

so that it is also similar; [VI. 4 and Def. 1] therefore the angle ABE is equal to the angle FGL.