: 3. the sector BGC be to the sector EHF. ence CK, the remaining part of the whole cir+3 Ax. cumference of the circle ABC, is equal t to the remaining part of the whole circumference of the same circle : therefore the angle BXC is * 27. 3. equal* to the angle COK; and the segment *11 Def. BXC is therefore similar to the segment COK;* and they are upon equal straight lines, BC, CK: but similar segment of circles upon equal * 24. 3. straight lines, are equal * to one another; there fore 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, whát multiple soever the circumference BL is of the circumference BC, the same multiple is the seetor 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 #5Def.5 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, which wise 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. Describe the circle * ACB about the triangle, * 5. 4. and produce AD to the circumference in E, and join EC: then because the angle BAD is equalt + Hyp. to the angle CAE, and the angle ABD to the angle * AEC, for they are in the same segment; the triangles B C DI ABD, AEC are equiangulart +32.1 to one another: Therefore as BA to AD, so is* EA to AC; E and consequently the rectan * 21. 3. #4. * * 3. 2. * 16.6. gle 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 * S5.3. is equal to the rectangle* BD, DC; therefore the rectangle BA, AC is equal to the rectangle BD, DC, together with the square of AD. Wherefore, if an angle, &c. Q. É. D. PROP. C. THEOR. 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 diumeter 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. Describe * the circle ACB about the triangle, and draw its diameter AE, and join FC: B because the right angle BDA is *51. $. equal* to the angle ECA in a semicircle, and the angle ABD E *21.3. equal * to the angle AEC in the same segment; the triangles to AD, so is EA to AC; and consequently the * 16.6. rectangle BA, AC is equal * to the rectangle EA, AD. If therefore from an angle, &c. *4.6. PROP. D. THEOR. The rectangle contained by the diagonals of a quadrilateral figure inscribed in a circle, is * equal to both the rectangles contained by its op- Let ABCD be any quadrilateral figureinscribed i terk in a circle, and join AC, BD: the rectangle cone recay tained by AC, BD shall be equal to the two ce of Al rectangles contained by AB, CD, and by AD, 3. BC. I Make the angle ABE equalf to the angle + 23. 1. EBC: and the angle BDA is equal* to the angle * 21. 3. - BCE, because they are in the same segment : let is therefore the triangle ABD is er en equiangular to the triangle B BCE: wherefore,* as BC is Get to CE, so is BD to DA; and consequently the rectangle BC, AD is equal * to the rect- E cause the angle ABE is equal is equal to the rectangle BD, AE: but the rectD'angle BC, AD has been shown equal to the rectangle BD,CE; therefore the whole rectangle AC, BD* is equal to the rectangle AB, *1. 2 DC, together with the rectangle AD, BC. Therefore the rectangle, &c. Q. E. D. * * 4. 6. * 16.6. This is a Lemma of Cl. Ptolomæus, in page 9. of his μεγαλη συνταξις. į Therefore the rectangles BC, AD and BA, DC are toge. ther equal + to the rectangles BD, CE and BD, AE; that is,+ 12 Ax. to the whole rectangle BD, AC. +1.2 214 THE ELEMENTS OF EUCLID. BOOK XI. DEFINITIONS. I. A solid is that which hath length, breadth, and thickness. II. a 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. |