Similarly it may be shewn that the sectors BGC, CGK, KGL are all equal ; and likewise the sectors EHF, FHM, MHN, NHR are all equal. .. the sector BGL is the same multiple of the sector BGC that the arc BL is of the arc BC; and the sector EHR is the same multiple of the sector EHF that the arc ER is of the arc EF. And if the arc BL= the arc ER, the sector BGL = the sector EHR: Proved. and if the arc BL is greater than the arc ER, the sector BGL is greater than the sector EHR: and if the arc BL is less than the arc ER, the sector BGL is less than the sector EHR. Now since there are four magnitudes, namely, the sectors BGC, EHF and the arcs BC, EF; and of the antecedents any equimultiples have been taken, namely the sector BGL and the arc BL; and of the consequents any equimultiples have been taken, namely the sector EHR and the arc ER: and since it has been shewn that the sector BGL is greater than, equal to, or less than the sector EHR, according as the arc BL is greater than, equal to, or less than the arc ER; .:. the four original magnitudes are proportionals; V. Def. 5. that is, the sector BGC : the sector EHF :: the arc BC : the arc EF. Q.E.D. : QUESTIONS FOR REVISION. 1. Explain why the operation known as Alternately requires that the four terms of a proportion should be of the same kind. Show that this is unnecessary in the case of Inversely. 2. State and prove algebraically the theorem known as Com. ponendo. In what proposition is this principle applied ? 3. Enunciate and prove algebraically the operation used in Book vi. under the name Ex Æquali. Also prove the same theorem in the following more general form: If there are two sets of magnitudes, such that the first is to the second of the first set as the first to the second of the other set, and the second to the third of the first set as the second to the third of the other, and so on throughout : then the first shall be to the last of the first set as the first to the last of the other. 4. Explain the operation Addendo, and give an algebraical proof of it. In what proposition of Book vi. is this operation employed ? 5. Give the geometrical and algebraical definitions of the ratio compounded of given ratios, and shew that the two definitions agree. By what artifice would Euclid represent the ratio compounded of the ratios A: B and C:D? 6. Two parallelograms ABCD, EFGH are equiangular to one another : if AB, BC are respectively 21 and 18 inches in length, and if EF, FG are 27 and 35 inches; shew that the areas of the parallelograms are in the ratio 2 : 5. : 7. If A : B=X : Y, and C : B=Z: Y; shew that A+C: B=X+2: Y. Explain and illustrate the necessity of the step invertendo in this proposition. 8. When is a straight line said to be divided in extreme and mean ratio? If a line 10 inches in length is so divided, shew that the lengths of the ents are approximately 6.2 inches and 3:8 inches. Shew also that the segments of any line divided in extreme and mean ratio are incommensurable. PROPOSITION B. THEOREM. If the vertical angle of a triangle be bisected by a straight line which cuts the base, the rectangle contained by the sides of the triangle shall be equal to the rectangle contained by the segments of the base, together with the square on the straight line which bisects the angle. E Let ABC be a triangle, having the BAC bisected by AD. Then shall the rect. BA, AC = the rect. BD, DC, with the sq. on AD. Describe a circle about the A ABC, IV. 5. and produce AD to meet the o ce in E. Join EC. Hyp. and the L ABD = the 2 AEC in the same segment; III. 21. .. the remaining - BDA = the remaining - ECA; I. 32. that is, the A BAD is equiangular to the A EAC. BA : AD :: EA : AC; .. the rect. BA, AC = the rect. EA, AD, VI. 16. = the rect. ED, DA, with the sq. on AD. VI. 4. II. 3. III. 35. But the rect. ED, DA = the rect. BD, DC; .. the rect. BA, AC = the rect. BD, DC, with the sq. on AD. Q.E.D. EXERCISE. If the vertical angle BAC is externally bisected by a straight line which meets the base in D, shew that the rectangle contained by BA, AC together with the square on AD is equal to the rectangle contained by the segments of the base, PROPOSITION C. THEOREM. If from the vertical angle of a triangle a straight line be drawn perpendicular to the base, the rectangle contained by the sides of the triangle shall be equal to the rectangle contained by the perpendicular and the diameter of the circle described about the triangle. Let ABC be a triangle, and let AD be the perp. from A to the base BC. Then the rect. BA, AC shall be equal to the rectangle contained by AD and the diameter of the circle circumscribed about the ДАВС. Describe a circle about the A ABC; IV. 5. draw the diameter AE, and join EC. Then in the A$ BAD, EAC, the rt. angle BDA= the rt. angle ECA, in the semicircle ECA, and the L ABD=the L AEC, in the same segment; III. 21. .. the remaining - BAD = the remaining _ EAC; 1. 32. that is, the A BAD is equiangular to the A EAC; .. BA : AD :: EA : AC; VI. 4. .. the rect. BA, AC = the rect. EA, AD. VI. 16. Q.E.D. PROPOSITION D. THEOREM. The rectangle contained by the diagonals of a quadrilateral inscribed in a circle is equal to the sum of the two rectangles contained by its opposite sides. с Let ABCD be a quadrilateral inscribed in a circle, and let AC, BD be its diagonals. Then the rect. AC, BD shall be equal to the sum of the rectangles AB, CD and BC, AD. Make the 2 DAE equal to the < BAC; I. 23. to each add the EAC, then the L DAC = the < EAB. Then in the AS EAB, DAC, the < EAB= the DAC, and the L ABE= the 2 ACD in the same segment; III. 21. .. the A* EAB, DAC are equiangular to one another; 1. 32. .. AB : BE :: AC : CD; VI. 4. ... the rect. AB, CD = the rect. AC, EB. VI. 16. Again in the A' DAE, CAB, the 2 DAE = the _ CAB, Constr. and the 2 ADE = the 2 ACB, in the same segment, III. 21. ... the A' DAE, CAB are equiangular to one another; 1. 32. .. AD : DE :: AC : CB; VI. 4. ... the rect. BC, AD = the rect. AC, DE. VI. 16. But the rect. AB, CD=the rect. AC, EB. Proved. .. the sum of the rects. BC, AD and AB, CD= the sum of the rects. AC, DE and AC, EB; that the sum of the BC, AD and AB, CD = the rect. AC, BD. II. 1. Q.E.D. |