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Join DE, and on the given straight line AB describe, as in the former case, the rectilineal figure ABHG, similar, and similarly situated to the rectilineal figure CDEF of four sides. At the point

B, in the straight line BH, make the angle HBL equal to the angle EDK; and at the point H, in the straight line BH, make the angle BHL equal to the

H

angle DEK;

[I. 23.

therefore the remaining angle at L is equal to the remaining angle at K.

Then, because the figures ABHG, CDEF are similar, the angle ABH is equal to the angle CDE; [VI. Def. 1. and the angle HBL is equal to the angle EDK;

[Constr. therefore the whole angle ABL is equal to the whole angle CDK.

[Axiom 2. For the same reason the whole angle GHL is equal to the whole angle FEK.

Therefore the five-sided figures ABLHG and CDKEF are equiangular to one another.

And, because the figures ABHG and CDEF are similar, therefore AB is to BH as CD is to DE; [VI. Definition 1. but BH is to BL as DE is to DK; [VI. 4. therefore, ex æquali, AB is to BL as CD is to DK. [V. 22. For the same reason, GH is to HL as FE is to EK. And BL is to LH as DK is to KE.

[VI. 4.

Therefore, the five-sided figures ABLHG and CDKEF are equiangular to one another, and have their sides about the equal angles proportionals;

therefore they are similar to one another. [VI. Definition 1.

In the same manner a rectilineal figure of six sides may be described on a given straight line, similar and similarly situated to a given rectilineal figure of six sides; and so on.

Q.E.F.

[blocks in formation]

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

Let ABC and DEF be similar triangles, having the angle B equal to the angle E, and let AB be to BC as DE is to EF, so that the

side BC is homologous to the side EF: the triangle ABC shall be to the triangle DEF in the duplicate ratio of BC to EF.

Take BG a third proportional to BC and EF, so that BC may be to EF as EF is to BG;

and join AG.

[VI. 11.

[Construction. [V. 11.

Then, because AB is to BC as DE is to EF, [Hypothesis. therefore, alternately, AB is to DE as BC is to EF; [V. 16. but BC is to EF as EF is to BG; therefore AB is to DE as EF is to BG; that is, the sides of the triangles ABG and DEF, about their equal angles, are reciprocally proportional; but triangles which have their sides about two equal angles reciprocally proportional are equal to one another, [VI. 15. therefore the triangle ABG is equal to the triangle DEF.

And, because BC is to EF as EF is to BG,

therefore BC has to BG the duplicate ratio of that which BC has to EF. [V. Definition 10. But the triangle ABC is to the triangle ABG as BC is to BG; [VI. 1. therefore the triangle ABC has to the triangle ABG the duplicate ratio of that which BC has to EF.

But the triangle ABG was shewn equal to the triangle DEF; therefore the triangle ABC has to the triangle DEF the duplicate ratio of that which BC has to EF.

Wherefore, similar triangles &c. Q.E.D.

[V. 7.

COROLLARY. From this it is manifest, that if three

straight lines be proportionals, as the first is to the third, so is any triangle described on the first to a similar and similarly described triangle on the second.

PROPOSITION 20. THEOREM.

Similar polygons may be divided into the same number of similar triangles, having the same ratio to one another that the polygons have; and the polygons are to one another in the duplicate ratio of their homologous sides.

Let ABCDE, FGHKL be similar polygons, and let AB be the side homologous to the side FG: the polygons ABCDE, FGHKL may be divided into the same number of similar triangles, of which each shall have to each the same ratio which the polygons have; and the polygon ABCDE shall be to the polygon FGHKL in the duplicate ratio of AB to FG.

Join BE, EC, GL, LH.

Then, because the polygon ABCDE is similar to the polygon FGHKL, [Hypothesis. the angle BAE is equal to the angle GFL, and BA is to AE as GF is to FL. [VI. Definition 1.

And, because the triangles ABE and FGL have one angle of the one equal to one angle of the other, and the sides

K

F

about these equal angles proportionals,

therefore the triangle ABE is equiangular to the triangle

FGL,

and therefore these triangles are similar;

therefore the angle ABE is equal to the angle FGL.

[VI. 6.

[VI. 4.

But, because the polygons are similar,

[Hypothesis.

therefore the whole angle ABC is equal to the whole angle FGH; [VI. Definition 1.

therefore the remaining angle EBC is equal to the remaining angle LGH.

[Axiom 3.

And, because the triangles ABE and FGL are similar, therefore EB is to BA as LG is to GF;

[Hypothesis.

and also, because the polygons are similar, therefore AB is to BC as FG is to GH; [VI. Definition 1.

therefore, ex æquali, EB is to BC as LG is to GH; [V. 22. that is, the sides about the equal angles EBC and LGH are proportionals ;

therefore the triangle EBC is equiangular to the triangle LGH;

and therefore these triangles are similar.

[VI. 6.

[VI. 4.

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

Therefore the similar polygons ABCDE, FGHKL may be divided into the same number of similar triangles.

Also these triangles shall have, each to each, the same ratio which the polygons have, the antecedents being ABE, EBC, ECD, and the consequents FGL, LGH, LÍK ; and the polygon ABCDE shall be to the polygon FGHKL in the duplicate ratio of AB to FG.

For, because the triangle ABE is similar to the triangle FGL,

therefore ABE is to FGL in the duplicate ratio of EB to LG.

[VI. 19. For the same reason the triangle EBC is to the triangle LGH in the duplicate ratio of EB to LG.

Therefore the triangle ABE is to the triangle FGL as the triangle EBC is to the triangle LGH. [V. 11. Again, because the triangle EBC is similar to the triangle LGH,

therefore EBC is to LGH in the duplicate ratio of EC to LH.

[VI. 19.

For the same reason the triangle ECD is to the triangle LHK in the duplicate ratio of EC to LH.

Therefore the triangle EBC is to the triangle LGH as the triangle ECD is to the triangle LHK.

[V. 11.

But it has been shewn that the triangle EBC is to the triangle LGH as the triangle ABE is to the triangle FGL. Therefore as the triangle ABE is to the triangle FGL, so is the triangle EBC to the triangle LGH, and the triangle ECD to the triangle LHK;

[V. 11. and therefore as one of the antecedents is to its consequent so are all the antecedents to all the consequents; [V. 12. that is, as the triangle ABE is to the triangle FGL so is the polygon ABCDE to the polygon FGHKL.

[VI. 19.

But the triangle ABE is to the triangle FGL in the duplicate ratio of the side AB to the homologous side FG; therefore the polygon ABCDE is to the polygon FGHKL in the duplicate ratio of the side AB to the homologous side FG.

Wherefore, similar polygons &C. Q.E.D.

COROLLARY 1. In like manner it may be shewn that similar four-sided figures, or figures of any number of sides, are to one another in the duplicate ratio of their homologous sides; and it has already been shewn for triangles; therefore universally, similar rectilineal figures are to one another in the duplicate ratio of their homologous sides.

COROLLARY 2. If to AB and FG, two of the homologous sides, a third proportional M be taken,

M

[VI. 11.

then AB has to M the duplicate ratio of that which AB has to FG.

[V. Definition 10.

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