Join DE. Then since when any number of magnitudes are proportional, as one antecedent is to its consequent, so are all the antecedents taken together to all the consequents together, .. AD : AE :: AB: AC, and DE is parallel to BC. Join KO, and let it meet DE in I. The triangles BOK, IOE are similar, and therefore, BK: KO:: EI: 10, and for the same reason, CK : KO :: DI: 10, whence EI=DI, and DE is bisected by KO; and it is also bisected by AK, :. AK passes through O. In the same manner it may be shewn that BG and CF, as also BI, CH intersect each other in points which are in the line AK. (34.) If from any point in one side of a triangle, two lines be drawn, one to the opposite angle, and the other parallel to the base, and the former intersect a line drawn from the vertex bisecting the base; this point of intersection, that of the line parallel to the base and the third side, and the third angular point are in the same straight line. From any point D in the side AB of the triangle ABC, let DE be drawn parallel to AC, and DC joined; and let DC meet BF drawn from B to the middle of AC in G; A, G, E are in the same straight line. B D E Let DE cut BF in K. The triangles DGK, CGF are equiangular, and .. DG : GC :: DK : FC :: DE : AC; hence the triangles DGE, AGC, having one angle in each equal, viz. EDG, GCA, and the sides about them proportional, are therefore similar; whence the angles AGC, DGE are equal; and DGC being a straight line, AGE is also. (35.) If one side of a triangle be divided into any two parts, and from the point of section two straight lines be drawn parallel to, and terminating at the other sides, and the points of termination be joined; and any other line be drawn parallel to either of the two former lines, so as to intersect the other, and to terminate in the sides of the triangle; then the two extreme parts of the three segments into which the line so drawn is divided will always be in the ratio of the segments of the first divided line. B M D N E G AK A F L Let AB be divided into any two parts in D, from which draw DE, DF parallel to the other two sides of the triangle; join EF, and draw GH parallel to DE, meeting DF and EF in I and K; GI: KH :: AD: DB; and if LM be parallel to DF, LK: MN :: AD : DB. Since GI is parallel to AF, and NK to DF, GI: AF :: (ID=) NK : FD :: NE : DE :: KH : FC, .. GI: KH :: AF: FC :: AD: DB, since DF is parallel to BC. Again since ML is parallel to BC, MN: BEND: DE: KF FE: KL EC, .. MN: KL :: BE: EC :: BD: DA. (36.) If through the point of bisection of the base of a triangle any line be drawn, intersecting one side of the triangle, and the other produced, and meeting a parallel to the base from the vertex; this line will be cut harmonically. From the vertex B of the triangle ABC, let BE be drawn parallel to the base AC, and through the middle point D let any line EGF be drawn meeting AB, BC, BE, in F, G, E; EG: DG :: FE: FD. Since AD is parallel to BE, F FE FD: BE: AD, : B but BE (DC=) DA :: EG: GD, since the triangles BGE, DGC are equiangular, .. (Eucl. v. 15.) EG: GD :: FE : FD, or the line is divided harmonically. (37.) If from either angle of a triangle a line be drawn intersecting that which joins the vertex and the bisection of the base, the opposite side, and the line from the vertex parallel to the base; it will be cut harmonically. From the vertex A of the tri- E angle ABC, let AE be drawn parallel to the base BC, and AD to its point of bisection D; and from C draw any line CFGE; then will CE CF:: EG: FG. D H Draw GH parallel to BC. Since AE and BC are parallel, (Eucl. vi. 2.) BA AG: CE: EG, and since GH is parallel to BD, BA: AG :: BD : GH :: DC : GH, (38.) To draw a line from one of the angles at the base of a triangle, so that the part of it cut off by a line drawn from the vertex parallel to the base, may have a given ratio to the part cut off by the opposite side. From A let AE be drawn parallel to BC. Divide AB in G, so that E G For the triangles AGE, BGC are equiangular, whence (Eucl. v. 18.) CE EG :: BA: AG, i. e. in the given ratio. (39.) To determine that point in the base produced of a right-angled triangle, from which the line drawn to the angle opposite to the base shall have the same ratio to the base produced, which the perpendicular has to the base itself. Let AB be the base, and CB the perpendicular of a right-angled triangle. Draw CE at right angles to AC, meeting AB produced in E. At the point C make the angle ECD=CAB. Dis the point required. D E B From D draw DF perpendicular to AC, and .. parallel to CE. Since the angle FDC is equal to the alternate angle DCE, i. e. to CAB, and the angles at F and B are right angles, .. the triangles DCF, ACB are equiangular; and DAF is also equiangular to ACB, hence FD DA: BC : CA, and DC: DF:: AC: AB .. ex æquo per. CD: DA :: CB BA. (40.) If the base of any triangle be divided into two parts by a line which is a mean proportional between them, and which being drawn parallel to the second side is terminated in the third; any line parallel to the base will be divided by the mean proportional (produced if necessary) into segments, which will be to each other inversely as the whole mean proportional to that segment which is terminated in the third side of the triangle. Let AC the base of the triangle ABC be divided into two parts in D, by a line DE which is parallel to BC, and a mean proportional between AD and DC; then any line FG parallel to AC, and meeting H E F G Н A D DE (produced if necessary) in H, will be divided into segments FH, HG, which are to each other inversely as the lines DE, HE. |