And Conversely, If the segments of the base produced have the same ratio, which the other sides of the triangle have to one another, the straight line drawn from the vertex to the point of section must bisect the exterior angle of the triangle. Let BD be to DC as BA is to AC, and join AD. Then must 4 CAD LEAD. For, the same construction being made, Нур. Ex. 3. If the base be divided into two segments, having the same ratio with the segments specified in the Proposition, the straight lines, drawn from the two points of section to the vertex of the triangle, are at right angles to each other. Ex. 4. If the angle, between the external bisector and a side, be equal to the angle, between the external bisector and the base, the perpendicular to the greater side, through the vertex, will bisect the segment of the base, cut off between the bisecting lines. PROPOSITION IV. THEOREM. The sides about the equal angles of triangles, which are equiangular to one another, are proportionals; and those which are opposite to the equal angles, are homologous sides. Let ABC, DEF be two As, having the ▲ s at A, B, C equal to the 4 s at D, E, F respectively. Then must the sides about the equal 48 be proportionals, those being homologous sides, which are opposite the equal ▲ s. For suppose ▲ DEF to be applied to ▲ ABC, Let G and H be the points in AB and AC, or these produced, on which E and F fall. Join GH. GH will be || to BC, ·.· 2 AGH= 4 ABC. I. 28. Then BA is to GA as CA is to HA, and .. BA is to ED as CA is to FD, whence BA is to AC as ED is to DF. VI. 2. V. 6. V. 15. Similarly, by applying the ▲ DEF, so that the 4s at F, E may coincide with those at C, B successively, we might show that AC is to CB as DF is to FE, and that CB is to BA as FE is to ED. Q. E. D. Ex. Divide a given angle into two parts, such that the perpendiculars from any point of the dividing line upon the *wo arms of the angle may be in a given ratio. PROPOSITION V. THEOREM. If the sides of two triangles, about each of their angles, be proportionals, the triangles must be equiangular to one another, and must have those angles equal, which are opposite to the homologous sides. Let the As ABC, DEF have their sides proportional, and AC is to CB as DF is to FE, Then must ▲ ABC be equiangular to ▲ EDF, those s being equal, which are opposite to the homologous sides, that is, ▲ BAC= ▲ EDF, and ▲ ABC= ¿ DEF, and ▲ ACB= 1 DFE. In AB, produced if necessary, make AG=DE, and draw GH || to BC, meeting AC in H. But ED is to DF as BA is to AC; So also it may be shown that GH=EF. GA-ED, and AH-DF, and HG=FE, I. 31. I. 29. Hyp. V. 5. V. 14. ..L GAH LEDF; LAGH=▲ DEF; ▲ AHG= L DFE. = I. c. But GAH ▲ BAC; ↳ AGH= 2 ABC; ▲ AHG= L ACB. ..L BAC=▲ EDF; 4 ABC= ▲ DEF, and ▲ ACB= 4 DFE. Q. E. D. PROPOSITION VI. THEOREM. If two triangles have one angle of the one equal to one angle of the other, and the sides about the equal angles proportionals, the triangles must be equiangular to one another, and must have those angles equal, which are opposite to the homologous sides. E In the As ABC, DEF, let ▲ BAC= L EDF, Then must ▲ ABC be equiangular to ▲ DEF, and ▲ ABC= ▲ DEF, and ▲ ACB= 2 DFE. L In AB, produced if necessary, make AG=DE, and draw GH || to BC. Then ▲ AGH is equiangular to ▲ ABC, and.. GA is to AH as BA is to AC, and.. GA is to AH as ED is to DF. But GA =ED, by construction, and .. AH=DF. I. 31. I. 29. VI. 4. V. 5. V. 14. I. 4. Q. E. D. Then ·.· GA =ED, and AH=DF and ▲ GAH= ▲ EDF'; :. 2 AGH= 2 DEF, and ▲ AHG= 2 DFE, and .. 4 ABC= ▲ DEF, and ▲ ACB= 2 DFE. Ex. 1. If from B, C, the extremities of the base of a triangle ABC, be drawn BD, CE, perpendicular to the opposite sides, shew that the triangles ADE, ABC are equiangular. Ex. 2. A variable chord OP is drawn through a fixed point O on the circumference of a circle, and Q is taken in it, so that the rectangle OP, OQ is constant, find the locus of Q. 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. |