at their circumferences: as the circumference BC to the circumference EF, so shall the angle BGC be to the angle EHF, and the angle BAC to the angle EDF; and also the sector BGC to the sector EHF.

Take any number of circumferences CK, KL, each equal to BC, and any number whatever FM, MN, each equal to EF: and join GK, GL, HM, HN. Because the circumferences BC, CK, KL, are all equal, the angles BGC, CGK, KGL are also all* equal: there- 27.3. fore what multiple soever the circumference BL is of the circumference BC, the same multiple is the angle BGL of the angle BGC: 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* 27.3. to the angle EHN; 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 ifequal, equal; and ifless, less: therefore as the circumference BC to the circumference EF, so is the angle BGC to the angle EHF:

D

H N

K

M

B

* 5 Def. 5.

15.5.

but as the angle BGC is to the angle EHF, so is the angle BAC to the angle EDF: for each is double * of * 20. 3. each; 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. Join BC, CK, and in the circumferences, BC, CK take any points X, O, 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

*4.1.

† 3 Ax.

* 27.3.

* 24.3.

*

*

equal angles; the base BC is equal to the base CK, 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: therefore the angle BXC is equal to the angle COK; and the segment BXC is therefore similar to the seg11 Def. 3. ment COK*; and they are upon equal straight lines, BC, CK: but similar segments of circles upon equal straight lines, are equal to one another; 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 A
be greater than EN, the sec-
tor BGL is greater than the

D

H

N

K

M

BX

E

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; * 5 Def. 5. 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. Wherefore, in equal circles, &c. Q. E. D.

PROP. B. THEOR.

If an angle of a triangle be bisected by a straight line, See N. which likewise cuts 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 of 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 of AD.

+ Hyp.

* 21. 3.

B

C

† 32. 1.

* 4. 6.

Describe the circle* ACB about the triangle, and * 5. 4. produce AD to the circumference in E, and join EC: then because the angle BAD is equalt to the angle CAE, and the angle ABD to the angle* AEC, for they are in the same segment; the triangles ABD, AEC are equiangular+ to one another: therefore as BA to AD, so is* EA to AC; and consequently the rectangle BA, AC is equal to the rectangle EA, AD, that is to the rectangle ED, DA, together with the square of AD: but the rectangle ED, DA is equal to the rectangle* BD, DC; there- * 35.3. fore the rectangle BA, AC is equal to the rectangle BD, DC, together with the square of AD. Wherefore, if an angle, &c. Q. E. D.

PROP. C. THEOR.

E

* 16. 6.

3.2.

If from any angle of a triangle a straight line be drawn See N. 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.

* 5. 4.

31. 3.

* 21. 3.

* 4. 6. * 16. 6.

See N.

† 23. 1.

21.3.

4. 6.

* 16. 6.

21.3.

1. 2.

B

Describe the circle ACB about the triangle, and draw its diameter AE, and join EC: because the right angle BDA is equal to the angle ECA in a semicircle, and the angle ABD equal* to the angle AEC in the same segment; the triangles ABD, AEC are equiangular: therefore as* BA to AD, so is EA to AC; and consequently the rectangle BA, AC is equal to the rectangle EA, AD. If therefore from an angle, &c.

Q. E. D.

PROP. D. THEOR.

E

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

B

Make the angle ABE equal to the angle DBC: add to each of these 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: therefore the triangle ABD is equiangular to the triangle BCE: wherefore*, as BC is to CE, so is BD to DA; and consequently the rectangle BC, AD, is equal* to the rectangle BD, CE: again, because the angle ABE is equal to the angle DBC, and the angle* BAE to the angle BDC, the triangle ABE is equiangular to the triangle BCD; therefore as BA to AE, so is BD to DC; wherefore the rectangle BA, DC is equal to the rectangle BD, AE: but the rectangle BC, AD has been shewn 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. Therefore the rectangle, &c. Q. E. D.

This is a Lemma of Cl. Ptolemæus, in page 9 of his Mayáλn Σύνταξις.

THE

ELEMENTS OF EUCLID.

BOOK XI.

DEFINITIONS.

I.

SOLID is that which hath length, breadth and thickness.

II.

That which bounds a solid is a superficies.

III.

A straight line is perpendicular, or at right angles, to a plane, when it makes right angles with every straight line in that plane which meets it.

IV.

A plane is perpendicular to a plane, when the straight lines drawn in one of the planes perpendicular to the common section of the two planes are perpendicular to the other plane.

V.

The inclination of a straight line to a plane, is the acute angle contained by that straight line, and another drawn from the point in which the first line meets the plane, to the point in which a perpendicular to the plane drawn from any point of the first line above the plane, meets the same plane.