Miscellaneous Exercises on Props. I. to VI. 1. If two triangles stand on the same base, and their vertices be joined by a straight line, the triangles are as the parts of this line intercepted between the vertices and the base. 2. If a circle be described on the radius of another circle as its diameter, and any straight line be drawn through the point of contact, cutting the two circles, the part, intercepted between the greater and lesser circles, shall be equal to the part within the lesser circle. 3. The side BC, of a triangle ABC, is bisected in D, and any straight line is drawn through D, meeting AB, AC, produced if necessary, in E, F, respectively, and the straight line through A, parallel to BC, in G. Prove that DE is to DF as GE is to GF. 4. If the angle A, of the triangle ABC, be bisected by AD, which cuts BC in D, and O be the middle point of BC, then OD bears the same ratio to OB that the difference of the sides bears to their sum. 5. The diameters of two circles and the distances between their centres are as the numbers 5, 4, 3; find the proportionate distances between the points of intersection of their common tangents. 6. If D, E be points in the sides AB, AC respectively of the triangle ABC, such that the triangles DAC, EAB are equal, shew that the sides AB, AC are divided proportionally in D and E. 7. If two of the exterior angles, of a triangle ABC, be bisected by the lines COE, BOD, intersecting in O, and meeting the opposite sides in E and D, prove that OD is to OB as AD is to DB, and that OC is to OE as AC is to AE. 8. B, C, the angles at the base of an isosceles triangle, are joined to the middle points, E, F, of AB, AC, by lines intersecting in C. Shew that the area BCG is equal to the area AEFG. 9. If, through any point in the diagonal of a parallelogram, a straight line be drawn, meeting two opposite sides of the figure, the segments of this line will have the same ratio as those of the diagonal. 10. The sides AB, AC, of a given triangle ABC, are produced to any points D and E, and the straight line DE is divided in F, so that DF is to FE as BD is to CE; shew that the locus of F is a straight line. PROPOSITION VII. THEOREM. If two triangles have one angle of the one equal to one angle of the other, and the sides about a second angle in each proportionals; then, if the third angles in each be both acute, both obtuse, or if one of them be a right angle, the triangles must be equiangular to one another, and must have those angles equal, about which the sides are proportionals. In the As ABC, DEF, let ▲ BAC= 2 EDF, and let AB be to BC as DE is to EF, and let s ACB, DFE be both acute, both obtuse, or let one of them be a right angle. Then must As ABC, DEF be equiangular to one another, having ABC= ▲ DEF, and ▲ ACB= ▲ DFE. For if ABC be not ▲ DEF, let one of them, as be greater than the other, and make ▲ ABG= ▲ DEF, and let BG meet AC in G. ABC, Then ▲ BAG= ▲ EDF, and ▲ ABG= 2 DEF, .. A ABG is equiangular to ▲ DEF, and .. AB is to BG as DE is to EF. But AB is to BC as DE is to EF, .. AB is to BG as AB is to BC, and.. BG=BC, and .. BCG= L BGC. I. 32. VI.4. Hyp. V. 5. V. 8. I. A. First, let then ACB and DFE be both acute, AGB is acute, and ..▲ BGC is obtuse ; I. 13. ..4 BCG is obtuse, which is contrary to the hypothesis. Next, let ACB and then DFE be both obtuse, AGB is obtuse, and .. ▲ BGC is acute; I. 13. .. [ BCG is acute, which is contrary to the hypothesis. Lastly, let one of the third 4 s ACB, DFE be a right ▲ . and.. 4s BCG, BGC together=two rt. 4 s, which is impossible. Hence ABC is not greater than ▲ DEF. So also we might shew that I. 17. DEF is not greater than N.B.-This Proposition is an extension of Proposition E of Book I. p. 42. Note.-We have made a slight change in Euclid's arrangement of the four Propositions that follow, because Eucl. vi. 8 is closely connected with the proof of Eucl. vi. 13. PROPOSITION VIII. PROBLEM. (Eucl. vi. 9.) From a given straight line to cut off any submultiple. B E Let AB be the given st. line. It is required to shew how to cut off any submultiple from AB. From A draw AC making any angle with AB. In AC take any pt. D, and make AC the same multiple of AD that AB is of the submultiple to be cut off from it. Join BC, and draw DE || to BC. Then ED is || to BC, .. CD is to DA as BE is to EA, and.. CA is to DA as BA is to EA. I. 31. VI. 2. V. 16. .. EA is the same submultiple of BA that DA is of CA. Hence from AB the submultiple required is cut off. Ex. 1. Cut off one-seventh of a given straight line. V. 19. Q. E. F. Note. This Proposition is a particular case of Proposition IX. PROPOSITION IX. PROBLEM. (Eucl. vi. 10.) To divide a given straight line similarly to a given straight line. Let AB be the st. line given to be divided, and AC the divided st. line. It is required to divide AB similarly to AC. Let AC be divided in the pts. D, E. Place AB, AC so as to contain any angle. Join BC, and through D, E draw DF, EG || to BC. I. 31. Ex. 1. Produce a given straight line, so that the whole produced line shall be to the produced part in a given ratio. Ex. 2. On a given base describe a triangle, with a given vertical angle and its sides in a given ratio. |