AD: And these figures are equiangular; therefore their sides about the equal angles are reciprocally proportional: (14. 6.) Wherefore, as FE to ED, so AE to EB: But FE is equal to AC, (34. 1.) that is to AB; and ED is equal to AE: Therefore as BA to AE, so is AE to AB: But AB is greater than AE; wherefore AE is greater than EB: (14.5.) Therefore the straight line AB is cut in extreme and mean ratio in E. (3. def. 6.) Which was to be done. B Otherwise, Let AB be the given straight line; it is required to cut it in extreme and mean ratio. Divide AB in the point C, so that the rectangle contained by AB, BC be equal to the square of AC: (11.2.) Then, because the rectangle AB, BC is equal A с в to the square of AC, as BA to AC so is AC to CB: (17. 6.) Therefore AB is cut in the extreme and mean ratio in C. (3. def. 6.) Which was to be done. PROP. XXXI. THEOREM. In right angled triangles, the rectilineal figure described upon the side opposite to the right angle, is equal to the similar and similarly described figures upon the sides containing the right angle. Let ABC be a right angled triangle, having the right angle BAC: The rectilineal figure described upon BC is equal to the similar, and similarly described figures upon BA, AC. Draw the perpendicular AD; there fore, because in the right angled triangle ABC, AD is drawn from the right angle at A perpendicular to the base BC, the triangles ABD, ADC are similar to the whole triangle ABC, and to one another, (8. 6.) and because the triangle ABC is similar to ABD, as CB to BA, so is BA to BD; (4. 6.) and because these three straight lines are proportionals, as the first to the third, so is the figure upon the first to the similar, and similarly described figure upon the second: (2 Cor. 20. 6.) Therefore as D C CB to BD, so is the figure upon CB to the similar and similarly described figure upon BA: And, inversely, (B. 5.) B as DB to BC, so is the figure upon BA to that upon BC: For the same reason, as DC to CB, so is the figure upon CA to that upon CB. Wherefore as BD and DC together to BC, so are the figures upon BA, AC to that upon BC: (24. 5.) But BD and DC together are equal to BC. Therefore the figure described on BC is equal (A. 5.) to the similar and similarly described figures on BA, AC. Wherefore, in right angled triangles, &c. Q. E. D. PROP. XXXII. THEOREM. If two triangles which have two sides of the one proportional to two sides of the other be joined at one angle, so as to have their homologous sides parallel to one another; the remaining sides shall be in a straight line. Let ABC, DCE be two triangles which have the two sides BA, AC proportional to the two CD, DE, viz. BA to AC, as CD to DE; and let AB be parallel to DC, and AC to DE, BC and CE are in a straight line. Because AB is parallel to DC, and the straight line AC A meets them, the alternate angles BAC ACD are equal; (29 1.) for the same rea son, the angle CDEB is equal to the angle ACD; wherefore also BAC is equal to CDE: And because the triangles ABC, DCE have one angle at A equal to one at D, and the sides about these angles proportionals, viz. BA to AC, as CD to DE, the triangle ABC is equiangular (6. 6.) to DCE: Therefore the angle ABC is equal to the angle DCE: And the angle BAC was proved to be equal to ACD: Therefore the whole angle ACE is equal to the two angles ABC, BAC; add the common angle ACB, then the angles ACE, ACB are equal to the angles ABC, BAC, ACB: But ABC, BAC, ACB are equal to two right angles; (32. 1.) therefore also the angles ACE, ACB are equal to two right angles: And since at the point C, in the straight line AC, the two straight lines BC, CE, which are on the opposite sides of it, make the adjacent angles ACE, ACB equal to two right angles; therefore (14. 1.) BC and CE are in a straight line. Wherefore, if two triangles, &c. Q. E. D. PROP. XXXIII, THEOREM. In equal circles, angles, whether at the centres or circumferences, have the same ratio which the circumferences on which they stand have to one another: So also have the sectors. Let ABC, DEF be equal circles; and at their centres the angles BGC, EHF, and the angles BAC, EDF at their circumferences; as their circum ference BC to the circumference EF, so is the angle BGC 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: (27. 3.) Therefore 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 As therefore the circumference BC to the circumference EF, so (5. def. 5.) is the angle BGC to the angle EHF: But as the angle BGC is to the angle EHF, so is (15. 5.) the angle BAC to the angle EDF; for each is double of each (20. 3.): 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 is the sector BGC 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, and that they contain equal angles; the base BC is equal (4. 1.) 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: Wherefore the angle BXC is equal to the angle COK (27. 3.); and the segment BXC is therefore similar to the segment COK (11. def. 3.); and they are upon equal straight lines BC, CK: But similar segments of circles upon equal straight lines, are equal (24. 3.) to one another: Therefore the segment BXC is equal to the segment COK; And the triangle BGC is equal to the triangle CGK; therefore the whole, the sector BCG, 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 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 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 that it has been proved, 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, (5. def. 5.) 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. THEOREM. If an angle of a triangle be bisected by a straight line 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 bisecting the angle. Let ABC be a triangle, and let the angle BAC be bisected by the straight line AD; the rectangle BA, AC is equal to the rectangle BD, DC, together with the square of AD. another: Therefore as BA to AD, so is (4. 6.) EA to AC, and consequently the rectangle BA, AC is equal (16.6.) to the rectangle EA, AD, that is, (3. 2.) to the rectangle ED, DA, together with the square of AD: But the rectangle ED, DA is equal to the rectangle (35.3.) 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. E. D PROP C. THEOREM. If from an 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. The rectangle contained by the diagonals of a quadrilateral inscribed in a circle, is equal to both the rectangles contained by its opposite sides. • This is a Lemma of C Ptolomæus, in page 9 of his μsyaλn ourraşı. equal (16.6.) to the rectangle BD, CE: Again, because the angle ABE is equal to the angle DBC, and the angle (21. 3.) BAE to the angle BDC, the triangle ABE is equiangular to the triangle BCD: As therefore 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 (4. 6.) is equal to the rectangle AB, DC, together with the rectangle AD, BC. Therefore the rectangle, &c. Q. E. D. I. BOOK XI. DEFINITIONS. A plane is perpendicular to a plane, when the straight lines drawn in one of the planes perpendicularly 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. VI. The inclination of a plane to a plane is the acute angle contained by two straight lines drawn from any the same point of their common section at right angles to it, one upon one plane, and the other upon the other plane. VII. Two planes are said to have the same, or a like inclination to one another, which two other planes have, when the said angles of inclination are equal to one another. VIII. Parallel planes are such which do not meet one another though produced. IX. A solid angle is that which is made by the meeting of more than two plane angles, which are not in the same plane, in one point. X. 'The tenth definition is omitted for reasons given in the notes.' XI. Similar solid figures are such as have all their solid angles equal each to each, and which are contained by the same number of similar planes. XII. A pyramid is a solid figure contained by planes that are constituted betwixt one plane and one point above it in which they meet. XIII. A prism is a solid figure contained by plane figures, of which two that are opposite are equal, similiar, and pa |