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Schol. 1.-This and the last propositions may be enunciated thus:-If the vertical angle of a triangle and its adjacent angle be bisected by lines cutting the base, it will be cut internally and externally in the ratio of the two sides.

Schol. 2.-By means of these two propositions, it is proved in optics, that the axis of a pencil of rays incident on a spherical mirror is divided harmonically by the radiant point, the geometrical focus of reflected rays, and the centre and surface of the reflector. It is also found that the lengths of three musical strings of the same thickness, material, and texture, and under the same tension, that produce any note, its fifth, and octave, are in harmonical progression; and hence the origin of the term. It is believed that Pythagoras first observed this relation of musical strings.

PROPOSITION IV. THEOREM.

The sides about the equal angles of equiangular triangles are proportionals; and those which are opposite to the equal angles are homologous sides, that is, are the antecedents or consequents of the ratios.

Let ABC, DCE, be equiangular triangles, having the angle ABC equal to the angle DCE, and the angle ACB to the angle DEC, and consequently the angle BAC equal to the angle CDE (I. 32). The sides about the equal angles of the triangles ABC, DCE, are proportionals; and those are the homologous sides which are opposite to the equal angles.

A

Let the triangle DCE be placed, so that its side CE may be contiguous to BC, and in the same straight line with it; and because the angles ABC, ACB, are together less than two right angles (I. 17), ABC and DEC, which is equal to ACB, are also less than two right angles; wherefore BA, ED, produced, shall meet (I. 29, Cor.); let them be produced and meet in the point F; and because the angle ABC is equal to the angle DCE, BF is parallel to CD (I. 28). Again, because the angle ACB is equal to the angle DEC, AC is parallel to FE; therefore FACD is a parallelogram; and consequently AF

is equal to CD, and AC to FD (I. 34); and because AC is parallel to FE, one of the sides of the triangle FBE, BA: AF:: BC: CE (VI. 2); but AF is equal to CD; therefore (V. 7), BA: CD:: BC: CE; and alternately, AB: BC: DC: CE. Again, because CD is parallel to BF, BC: CE:: FD: DE; but FD is equal to AC; therefore, BC: CE:: AC: DE; and alternately, BC: CA::CE: ED. Therefore, because it has been proved that AB: BC:: DC: CE; and BC: CA::CE: ED, by equality, BA: AC::CD: DE.

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If the sides of two triangles, about each of their angles, be proportionals, the triangles shall be equiangular, and have their equal angles opposite to the homologous sides.

Let the triangles ABC, DEF, have their sides proportionals, so that AB is to BC, as DE to EF; and BĈ to CA, as EF to FD; and consequently, by equality, BA to AC, as ED to DF; the triangle ABC is equiangular to the triangle DEF, and their equal angles are opposite to the homologous sides; namely, the angle ABC being equal to the angle DEF, and BCA to EFD, and also BAC to EDF.

E

F

At the points E, F, in the straight line EF, make (I. 23) the angle FEG equal to the angle ABC, and the angle EFG equal to BCA; wherefore the remaining angle BAC is equal to the remaining angle EGF (I. 32), and the triangle ABC is therefore equiangular to the triangle GEF; and consequently they have their sides opposite to the equal angles proportionals (VI. 4); wherefore, as AB to BC, so is GE to EF; but as AB to BC, so is DE to EF; therefore, as DE to EF, so GE to EF (V. 11); therefore DE and GE have the same ratio to EF, and consequently are equal (V. 9); for the same reason, DF is equal to FG; and because, in the triangles DEF, GEF, DE is equal to EG, and EF common, and also the base DF equal to the base GF; therefore (I. 8) the angle DEF is equal to the angle GEF, DFE to GFE, and EDF to EGF; and because the angle DEF is equal to the angle GEF, and GEF to the angle ABC; therefore the

angle ABC is equal to the angle DEF; for the same reason, the angle ACB is equal to the angle DFE, and the angle at A to the angle at D; therefore the triangle ABC is equiangular to the triangle DEF.

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If two triangles have one angle of the one equal to one angle of the other, and the sides about the equal angles proportionals, the triangles shall be equiangular, and shall have those angles equal which are opposite to the homologous sides.

Let the triangles ABC, DEF, have the angle BAC in the one equal to the angle EDF in the other, and the sides about those angles proportionals; that is, BA to AC, as ED to DF; the triangles ABC, DEF, are equiangular, and have the angle ABC equal to the angle DEF, and ACB to DFE.

At the points D, F, in the straight line DF, make (I. 23) the angle FDG equal to either of the angles BAC, EDF; and the angle DFG equal to the angle ACB; wherefore the remaining angle at B is equal to the A remaining one at G (I. 32), and consequently the triangle ABC is equiangular to the triangle DGF; and therefore as BA to AC, so is GD to DF (VI. 4); в

C E

D

but, by the hypothesis, as BA to AC, so is ED to DF; as therefore ED to DF, so is GD to DF (V. 11); wherefore ED is equal to DG (V. 9); and DF is common to the two triangles EDF, GDF; therefore the two sides ED, DF, are equal to the two sides GD, DF; but the angle EDF is also equal to the angle GDF; therefore (I. 4) the angle DFG is equal to the angle DFE, and the angle at G to the angle at E. But the angle DFG is equal to the angle ACB; therefore the angle ACB is equal to the angle DFE; and the angle BAC is equal to the angle EDF (Hyp.); wherefore also the remaining angle at B is equal to the remain ing angle at E; therefore the triangle ABC is equiangular to the triangle DEF.

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If two triangles have one angle of the one equal to one angle of the other, and the sides about two other angles proportionals, then, if each of the remaining angles be either less, or not less, than a right angle, the triangles shall be equiangular, and have those angles equal about which the sides are proportionals.

Let the two triangles ABC, DEF, have one angle in the one equal to one angle in the other; namely, the angle BAC to the angle EDF, and the sides about two other angles ABC, DEF, proportionals, so that AB is to BC, as DE to EF; and in the first case, let each of the remaining angles at C, F, be less than a right angle. The triangle ABC is equiangular to the triangle DEF; that is, the angle ABC is equal to the angle DEF, and the remaining angle at C to the remaining angle at F.

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For, if the angles ABC, DEF, be not equal, one of them is greater than the other. Let ABC be the greater, and at the point B, in the straight line AB, make the angle ABG equal to the angle DEF (I. 23); and because the angle at A is equal to the angle at D, and the angle ABG to the angle DEF; the remaining angle AGB is equal to the remaining angle DFE (I. 32); therefore the triangle ABG is equiangular to the triangle DEF; wherefore (VI. 4), as AB is to BG, so is DE to EF; but as DE to EF, so, by hypothesis, is AB to BC; therefore, as AB to BC, so is AB to BG (V. 11); and because AB has the same ratio to each of the lines BC, BG; BC is equal to BG (V. 9), and therefore the angle BGC is equal to the angle BCG (I. 6); but the angle BCG is, by hypothesis, less than a right angle; therefore also the angle BGC is less than a right angle, and the adjacent angle AGB must be greater than a right angle (I. 13). But it was proved that the angle AGB is equal to the angle at F; therefore the angle at F is greater than a right angle; but, by the hypothesis, it is less than a right angle; which is absurd; therefore the angles ABC, DEF, are not unequal, that is, they are equal; and the angle at

A is equal to the angle at D; wherefore the remaining angle at C is equal to the remaining angle at F; therefore the triangle ABC is equiangular to the triangle DEF.

Next, let each of the angles at C, F, be not less than a right angle; the triangle ABC is also in this case equiangular to the triangle DEF.

The same construction being made, it may be proved in like manner that BC is equal to BG, and the angle at C equal to the angle BGC;/ but the angle at C is not less than a right angle; therefore the angle BGC is not

F

less than a right angle; wherefore, two angles of the triangle BGC are together not less than two right angles, which is impossible (I. 17); and therefore the triangle ABC may be proved to be equiangular to the triangle DEF, as in the first case.

PROPOSITION VIII. THEOREM.

In a right-angled triangle, if a perpendicular be drawn from the right angle to the base, the triangles on each side of it are similar to the whole triangle, and to one another.

Let ABC be a right-angled triangle, having the right angle BAC; and from the point A let AD be drawn perpendicular to the base BC; the triangles ABD, ADC, are similar to the whole triangle ABC, and to one another.

B

D C

Because the angle BAC is equal to the angle ADB, each of them being a right angle, and that the angle at B is common to the two triangles ABC, ABD; the remaining angle ACB is equal to the remaining angle BAD (I. 32); therefore the triangle ABC is equiangular to the triangle ABD, and the sides about their equal angles are proportionals (VI. 4); wherefore the triangles are similar (VI. Def. 9); in the same manner, it may be demonstrated that the triangle ADC is equiangular and similar to the triangle ABC; and the triangles ABD, ADC, being both equiangular and similar to ABC, are equiangular and similar to each other.

COR. From this it is manifest that the perpendicular drawn from the right angle of a right-angled triangle

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