PD, PE; and from A draw AF, AG respectively parallel to them. Join PF, PG, AP. Those three lines will divide the triangle into three equal parts. F D Join AD, AE. Since AF is parallel to PD, the triangle APF is equal to ADF; to each of these add ABF, :. APFB is equal to ADB. In the same manner APGC is equal to AEC; and .. the remainder FPG is equal to DAE. Now the triangles ABD, ADE, AEC, being on equal bases and of the same altitude, are equal, .. APFB, PFG, APGC are also equal; and the triangle ABC is trisected. (19.) From a given point in the side of a triangle, to draw lines, which will divide the triangle into parts which shall have a given ratio. Let ABC be the given triangle, and P the given point in the side BC. Divide BC, in the points D, E, F, into parts which shall have the given ratio. Join AD, H AE, AF, AP; and draw DG, EH, FI parallel to AP. Join PG, PH, PI; they will divide the triangle, as required. For the triangles ABD, ADE, AEF, AFC being as their bases will be in the given ratio. And since DG is parallel to AP, the triangles DGA, DGP are equal, .. DBA, GPB are equal. And since the triangle ADP AGP, and AEP – AHP, .. ADE=HPG. Also APE=AHP, and APF-AIP, .. AEF=AHPI, and = = AFC=PIC; .. the parts PBG, GPH, HPIA, IPC are equal to ABD, ADE, AEF, AFC, and are .. in the given ratio. The same may be proved whatever be the number of parts. (20.) If two exterior angles of a triangle be bisected, and from the point of intersection of the bisecting lines, a line be drawn to the opposite angle of the triangle; it will bisect that angle. Let the exterior angles EBC, BCF, of the triangle ABC, be bisected by the lines BD, CD meeting in D. Join DA; it will bisect the angle BAC. Let fall the perpendiculars DE, DF, DG. Then the angles DBE, A B DBG being equal, and the angles at E and G being right angles, and DB common to the triangles DBE, DBG, .. DE=DG. In the same manner DG=DF; and .. DE=DF. Hence in the right-angled triangles DAE, DAF, DE is equal to DF and DA is common, .. the triangles are equiangular, and the angles DAE DAF are equal, i. e. BAC is bisected by AD. (21.) If in two triangles the vertical angle of the one be equal to that of the other, and one other angle of the former be equal to the exterior angle at the base of the latter; the sides about the third angle of the former shall be proportional to those about the interior angle at the base of the latter. Let ABC, DEF be two triangles having the angle BAC equal to EDF, and ABC equal to the exterior angle DFG, made by producing the side EF; then AC CB :: DE : EF. At the point D in the line FD, make the angle FDG equal to the angle EDF or BAC, and meeting EF pro duced in G. Since the angle FDG is equal to the angle BAC, and DFG is equal to ABC, .. the triangles ABC, DFG are equiangular, and AC: CB:: DG GF. But since the angle GDE is bisected by DF, .. (Eucl. vi. 3.) DG: GF:: DE EF .. AC: CB :: DE : EF. (22.) In a given triangle to draw a line parallel to one of the sides, so that it may be a mean proportional between the segments of the base. Let ABC be the given triangle; in the base of which take a point E, such that AE may be to EC in the duplicate ratio of AC: CB; draw ED parallel to BC; ED is the line required. A B E C Since ED is parallel to BC, AE : ED :: AC : CB. But AE EC in the duplicate ratio of AC: CB, and therefore also in the duplicate ratio of AE: ED; whence from the definition of the duplicate ratio, AE ED: ED: EC. (23.) To draw a line parallel to the common base of two triangles which have different altitudes, so that the parts of it intercepted by the sides may have a given ratio. A Let ABC, DBC be two triangles on the same base BC, the vertex D being in the side AC. Divide BC in E, so that BC: CE may be equal to the given ratio. Join AE, cutting BD in G; and through G draw FH parallel to BC; FH is the line required. Since FH is parallel to BC, FH : GH:: BC: CE, i. e. in the given ratio. F G I K LH But if the vertex I is not in AC, draw ID parallel to BC; join BD; divide the base BC, as before; join AE, and draw FK parallel to BC. Then it is evident that GH=LK, and ... FH: LK in the given ratio. COR. If the triangles be upon equal bases, but in the same straight line, the line may be drawn in a similar manner. (24.) If the base of a triangle be produced, so that the whole may be to the part produced in the duplicate ratio of the sides; the line joining the vertex and the extremity of the part produced will be a mean proportional between the whole line produced and the part produced. Let AC be produced to D, so that AD may be to DC in the duplicate ratio of AB: BC; join BD; it will be a mean proportional between AD and DC. Ꭺ B Draw CE parallel to AB; then AB : CE :: AD : DC, i. e. in the duplicate ratio of AB: BC, whence AB BC: BC: CE, i. e. the sides about the equal angles ABC, BCE are proportional; therefore the triangles ABC, BCE are similar, and the angle at A is equal to the angle CBD; .. the triangles ABD, CBD are equiangular, and AB: BD :: BD : DC. (25.) To determine a point within a given triangle, which will divide a line parallel to the base into two segments, such that the excess of each segment above the perpendicular distance between the parallel lines may be to each other in the duplicate ratio of the respective segments. Let ABC be the given triangle. From C draw CD perpendicular to AB, and from D draw DE, DF bisecting the angles ADC, BDC. Join BE, cutting CD in P; P is the point required. E PL K Through P draw GHIK parallel to AB; then the angle PDH is equal to the angle HDA, i. e. to the alternate angle PHD; and .. HP, and in like manner PI will each be equal to PD the perpendicular distance of GK from AB; and GH, IK will be equal to the ex |