And Conversely, If the similar figures, similarly described on four straight lines, be proportionals, those straight lines must be proportionals. The same construction being made, let KAB be to LCD as MF is to NH, then must AB be to CD as EF is to GH. Make as AB to CD so EF to PR, VI. 11. and on PR describe the rectilinear figure SR, similar and simi larly situated to either of the figures MF, NH. Then, by the first part of the proposition, KAB is to LCD as MF is to SR. But KAB is to LCD as MF is to NH. .. SR=NH, VI. 18. Hyp. V. 8. Also, SR and NH are similar and similarly situated, PROPOSITION XXV. THEOREM. (Eucl. vi. 33.) In equal circles, angles, whether at the centres or the circumferences, have to one another the same ratio as the arcs which subtend them; and so also have the sectors. M N In the equal os ABC, DEF let the ▲ 8 BGC, EHF at the centres, and the ▲ 8 BAC, EDF at the circumferences, be subtended by the arcs BC, EF. Then I. ▲ BGC must be to ▲ EHF as arc BC is to arc EF. Take any number of arcs CK, KL, each=BC, and any number of arcs FM, MN, NR each=EF. Then arcs BC, CK, KL are all equal, .. 4s BGC, CGK, KGL are all equal. III. 27. :. ▲ BGL is the same multiple of ▲ BGC that arc BL is of arc BC. So also, ▲ EHR is the same multiple of ▲ EHF that arc ER is of arc EF. And BGL is equal to, greater than, or less than ▲ EHR, according as arc BL is equal to, greater than, or less than .. BGC is to EHF as arc BC is to arc EF. V. Def. 5. II. L BAC must be to ▲ EDF as arc BC is to arc EF. For BGC=twice ▲ BAC, and ▲ EHF=twice ▲ EDF, ..LBAC is to 4 EDF as 4 BGC is to 4 EHF, III. 20. V. 11. and .. BAC is to ▲ EDF as arc BC is to arc EF. V. 5. III. Sector BGC must be to sector EHF as arc BC is to arc EF. III. 26, Cor. For sectors BGC, CGK, KGL are all equal, III. 26, Cor. .. sector BGL is the same multiple of sector BGC that arc BL is of arc BC, and sector EHR is the same multiple of sector EHF that arc ER is of arc EF; also, sector BGL is equal to, greater than or less than sector EHR, according as arc BL is equal to, greater than, or less than arc ER, III. 26. and.. sector BGC is to sector EHF as arc BC is to arc EF. Q. E. D. COR. In the same circle, angles, whether at the centres or the circumferences, have the same ratio as the arcs which subtend them; and so also have the sectors. 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 line bisecting the angle. Let BAC of the ▲ ABC be bisected by the st. line AD. A Then rect. BA, AC=rect. BD, DC together with sq. on AD. Describe the ABC about the ▲, produce AD to meet the Oce in E, and join EC. III. B. p. 135. Then BAD = L CAE, Hyp. and ABD = AEC, in the same segment, III. 21. 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. Ꭰ E Let ABC be a▲, and AD the 1 from A to BC. draw the diameter AE, and join EC. III. B. III. 31. III. 21. I. 32. VI. 4. VI. 16. Q. E. D. Ex. 1. Shew that the rectangle contained by the two sides can never be less than twice the triangle. Ex. 2. ABC is a triangle, and AM the perpendicular upon BC, and P any point in BC; if O, O' be the centres of the circles described about ABP, ACP, the rectangle AP, BC is double of the rectangle of AM, 00'. Ex. 3. A bisector of an angle of a triangle is produced to meet the circumscribed circle. Prove that the rectangle, contained by this whole line and the part of it within the triangle, is equal to the rectangle contained by the two sides. |