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. B с Then BC and CE shall be in a straight line. Demonstration. Because AB is parallel to DC, and the straight line AC meets them, E 1. The alternate angles BAC, ACD are equal (I. 29); for the same reason, the angle CDE is equal to the angle ACD; wherefore also 2. BAC is equal to CDE (Ax. 1); 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 to DCE (VI. 6); therefore 3. The angle ABC is equal to the angle DCE; and the angle BAC was proved to be equal to ACD; therefore 4. The whole angle ACE is equal to the two angles ABC, BAC (Ax. 2); add to each of these equals the common angle ACB, then 5. The angles ACE, ACB are equal to the angles ABC, BAC, ACB; but ABC, BAC, ACB are equal to two right angles (I. 32); therefore also 6. 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 7. BC and CE are in a straight line (I. 14). Wherefore, if two triangles, &c. Q.E.D. PROPOSITION 33.-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. A Then 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. с K E for the same reason, Construction. 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. Demonstration. Because the circumferences BC, CK, KL are all equal, the angles BGC, CGK, KGL are also all equal (III. 27) ; therefore 1. What multiple soever the circumference BL is of the circumference BC, the same multiple is the angle BGL of the angle BGC; 2. 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 (III. 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 ÊHN; 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 3. 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 4. 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. K E F M B X Construction. Join BC, CK, and in the circumferences BC, CK, take any points X, O, and join BX, XC, CO, OK. Demonstration. 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; 1. The base BC is equal to the base CK (I. 4), and the triangle GBC to the triangle GCK; and because the circumference BC is equal to the circumference CK, 2. 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 (III. 27); and therefore 3. The segment BXC is 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 4. The segment BXC is equal to the segment COK; and the triangle BGC was proved to be equal to the triangle CGK; therefore 5. The whole, the sector BGC, is equal to the whole, the sector CGK; for the same reason, 6. 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 7. 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, 8. 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 9. 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 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 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. Then the rectangle BA, AC shall be equal to the rectangle BD, DC, together with the square on AD. D Construction. Describe the circle ACB about the triangle (IV. 5), and produce AD to the circumference in E, and join EC. Demonstration. Because the angle BAD is equal to the angle CAE (hyp.), and the angle ABD to the angle A EC (III. 21), for they are in the same segment; the triangles ABD, AEC ar equiangular to one another (I. 32); therefore 1. As BA to AD, so is EA to AC (VI. 4); and consequently the rectangle BA, AC is equal to the rectangle EA, AD (VI. 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 2. 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.-Theorem. 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. Then the rectangle BA, AC shall be equal to the rectangle contained by AD and the diameter of the circle described about the triangle. Construction. Describe the circle ACB about the triangle (IV. 5), and draw its diameter AE, and join EC. |