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and join GK, GL, HM, HN.

Because the circumferences BC, CK, KL are all equal,

the angles BGC, CGK, KGL are also all equal: (III. 27.) therefore what multiple soever the circumference BL is of the circumference BC, the same multiple is the angle BGL of the angleBGC:

for the same reason, whatever multiple the circumference EN is of the circumference EF, the same multiple is the angle EHN of the angle EHF:

and if the circumference BL be equal to the circumference EN, the angle BGL is also equal to the angle EHN; (111. 27.)

and if the circumference BL be greater than EN,

likewise the angle BGL is greater than EHN; and if less, less; therefore, since there are four magnitudes, the two circumferences BC, EF, and the two angles BGC, EHF; and that of the circumference BC, and of the angle BGC, have been taken any equimultiples whatever, viz. the circumference BL, and the angle BGL; and of the circumference EF, and of the angle EHF, any equimultiples whatever, viz. the circumference EN, and the angle EHN:

and since it has been proved, that if the circumference BL be greater than EN;

the angle BGL is greater than EHN;

and if equal, equal; and if less, less;

therefore as the circumference BC to the circumference EF, so is the angle BGC to the angle EHF: (v. def. 5.)

but as the angle BGC is to the angle EHF, so is the angle BAC to the angle EDF: (v. 15.)

for each is double of each; (III. 20.)

therefore, as the circumference BC is to EF, so is the angle BGC to the angle EHF, and the angle BAC to the angle EDF.

Also, as the circumference BC to EF, so shall the sector BGC be to the sector EHF.

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Join BC, CK, and in the circumferences, BC, CK, take any poinst X, 0, and join BX, XC, CO, OK.

Then, because in the triangles GBC, GCK,

the two sides BG, GC are equal to the two CG, GK each to each, and that they contain equal angles;

the base BC is equal to the base CK, (1. 4.)

and the triangle GBC to the triangle GCK:

and because the circumference BC is equal to the circumference CK, the remaining part of the whole circumference of the circle ABC, is equal to the remaining part of the whole circumference of the same circle: (ax. 3.)

therefore the angle BXC is equal to the angle COK; (1. 27.) and the segment BXC is therefore similar to the segment COK

(III. def. 11.)

and they are upon equal straight lines, BC, CK:

but similar segments of circles upon equal straight lines, are equal to one another: (III. 24.)

therefore the segment BXC is equal to the segment COK: and the triangle BGC was proved to be equal to the triangle CGK; therefore the whole, the sector BGC, is equal to the whole, the sector CGK:

for the same reason, the sector KGL is equal to each of the sectors BGC, CGK:

in the same manner, the sectors EHF, FHM, MHN may be proved equal to one another:

therefore, what multiple soever the circumference BL is of the circumference BC, the same multiple is the sector BGL of the sector BGC; and for the same reason, whatever multiple the circumference EN is of EF, the same multiple is the sector EHN of the sector EHF:

and if the circumference BL be equal to EN, the sector BGL is equal to the sector EHN;

and if the circumference BL be greater than EN, the sector BGL is greater than the sector EHN;

and if less, less;

since then, there are four magnitudes, the two circumferences BC, EF, and the two sectors BGC, EHF, and that of the circumference BC, and sector BGC, the circumference BL and sector BGL are any equimultiples whatever; and of the circumference EF, and sector EHF, the circumference EN, and sector EHN are any equimultiples whatever;

and since it has been proved, that if the circumference BL be greater than EN, the sector BGL is greater than the sector EHN; and if equal, equal; and if less, less:

therefore, as the circumference BC is to the circumference EF, so is the sector BGC to the sector EHF. (v. def. 5.)

Wherefore, in equal circles, &c.

Q. E. D.

PROPOSITION B. THEOREM.

If an angle of a triangle be bisected by a straight line which likewise outs the base; the rectangle contained by the sides of the triangle is equal to the rectangle contained by the segments of the base, together with the square on the straight line which bisects the angle.

Let ABC be a triangle, and let the angle BAC be bisected by the straight line AD.

The rectangle BA, AC shall be equal to the rectangle BD, DC, together with the square on AD.

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Describe the circle ACB about the triangle, (IV. 5.) and produce AD to the circumference in E, and join ÉC. Then because the angle BAD is equal to the angle CAE, (hyp.) and the angle ABD to the angle AEC, (III. 21.)

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for they are in the same segment;

the triangles ABD, AEC are equiangular to one another: (1. 32.) therefore as BA to AD, so is EA to AC; (VI. 4.)

and consequently the rectangle BA, AC

is equal to the rectangle EA, AD, (yı. 16.) that is, to the rectangle ED, DA, together with the square on AD; (II. 3.)

but the rectangle ED, DA is equal to the rectangle BD, DC; (III. 35.) therefore the rectangle BA, AC is equal to the rectangle BD, DC, together with the square on AD. Wherefore, if an angle, &c.

Q.E.D.

PROPOSITION C. THEQREM.

If from any angle of a triangle, a straight line be drawn perpendicular to the base; the rectangle contained by the sides of the triangle is equal to the rectangle contained by the perpendicular and the diameter of the circle described about the triangle.

Let ABC be a triangle, and AD the perpendicular from the angle A to the base BC.

The rectangle BA, AC shall be equal to the rectangle contained by AD and the diameter of the circle described about the triangle.

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Describe the circle ACB about the triangle, (IV. 5.)

and draw its diameter AE, and join EC.

Because the right angle BDA is equal to the angle ECA in a semicircle, (III. 31.)

and the angle ABD equal to the angle AEC in the same segment; (III. 21.)

the triangles ABD, AEC are equiangular:

therefore as BA to AD, so is EA to AČ; (VI. 4.)

and consequently the rectangle BA, AC

is equal to the rectangle EA, AD. (vi. 16.)
If therefore, from any angle, &c.

Q. E.D.

PROPOSITION D. THEOREM.

The rectangle contained by the diagonals of a quadrilateral figure inscribed in a circle, is equal to both the rectangles contained by its opposite sides.

Let ABCD be any quadrilateral figure inscribed in a circle, and join AC, BD.

The rectangle contained by AC, BD shall be equal to the two rectangles contained by AB, CD, and by AD, BC.

Make the angle ABE equal to the angle DBC: (1. 23.)

add to each of these equals the common angle EBD,
then the angle ABD is equal to the angle EBC:

and the angle BDA is equal to the angle BCE, because they are in the same segment; (III. 21.)

therefore the triangle ABD is equiangular to the triangle BCE: wherefore, as BC is to CE, so is BD to DA; (VI. 4.)

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and consequently the rectangle BC, AD is equal to the rectangle BD, CE: (vi. 16.)

again, because the angle ABE is equal to the angle DBC, and the angle BAE to the angle BDC, (ш. 21.)

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the triangle ABE is equiangular to the triangle BCD:

therefore as BA to AE, so is BD to DC;

wherefore the rectangle BA, DO is equal to the rectangle BD, AE:
but the rectangle BC, AD has been shown to be equal
to the rectangle BD, CE;

therefore the whole rectangle AC, BD is equal to the rectangle
AB, DC, together with the rectangle AD, BC. (¤. 1.)

Therefore the rectangle, &c. Q. E. D.

This is a Lemma of Cl. Ptolemæus, in page 9 of his Mɛyáλŋ ZúvtαEIS..

NOTES TO BOOK VI.

In this Book, the theory of proportion exhibited in the Fifth Book, is applied to the comparison of the sides and areas of plane rectilineal figures, both of those which are similar, and of those which are not similar.

Def. I. In defining similar triangles, one condition is sufficient, namely, that similar triangles are those which have their three angles respectively equal; as in Prop. 4, Book vi, it is proved that the sides about the equal angles of equiangular triangles are proportionals. But in defining similar figures of more than three sides, both of the conditions stated in Def. 1, are requisite, as it is obvious, for instance, in the case of a square and a rectangle, which have their angles respectively equal, but have not their sides about their equal angles proportionals.

The following definition has been proposed: "Similar rectilineal figures of more than three sides, are those which may be divided into the same number of similar triangles." This definition would, if adopted, require the omission of a part of Prop. 20, Book vi.

Def. III.

To this definition may be added the following:

A straight line is said to be divided harmonically, when it is divided into three parts, such that the whole line is to one of the extreme segments, as the other extreme segment is to the middle part. Three lines are in harmonical proportion, when the first is to the third, as the difference between the first and second, is to the difference between the second and third; and the second is called a harmonic mean between the first and third.

The expression 'harmonical proportion' is derived from the following fact in the Science of Acoustics, that three musical strings of the same material, thickness and tension, when divided in the manner stated in the definition, or numerically as 6, 4, and 3, produce a certain musical note, its fifth, and its octave.

Def. IV. The term altitude, as applied to the same triangles and parallelograms, will be different according to the sides which may be assumed as the base, unless they are equilateral.

Prop. 1. In the same manner may be proved, that triangles and parallelograms upon equal bases, are to one another as their altitudes.

Prop. II. Observe, that the three diagrams include all the possible cases of this important proposition. In the first diagram, the line intersects the sides between the base and the vertex of the triangle: in the second, the line intersects the sides produced below the base: in the third, the line intersects the sides produced beyond the vertex.

Prop. A. When the triangle ABC is isosceles, the line which bisects the exterior angle at the vertex is parallel to the base. In all other cases,

if the line which bisects the angle BAC cut the base BC in the point G,
then the straight line BD is harmonically divided in the points G, C.
For BG is to GC as BA is to AC; (vi. 3.)
and BD is to DC as BA is to AC, (vI. A.)
therefore BD is to DC as BG is to GC,

but BG = BD – DG, and GC GD DC.

Wherefore BD is to DC as BD - DG is to GD - DC.

Hence BD, DG, DC, are in harmonical proportion.

Prop. IV is the first case of similar triangles, and corresponds to the third case

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