the lines which they represent are equal and parallel; i.e. CA is equal and parallel to BD. Ex. 2. The opposite sides of a parallelogram are equal; and the diagonals bisect each other. Since AB is parallel to CD, if vector AB be represented by a, vector CD will be represented by some numerical multiple of a (Art. 3), call it ma. And since CA is parallel to DB; if vector CA be ẞ, then vector DB is nẞ; hence Hence (Art. 6) m = 1, n = 1, i.e. the opposite sides of the parallelogram are equal. Again, as vectors, 40+0B = AB = CD = CO + OD; And as 40 is a vector along OD, and CO a vector along OB; it follows (Art. 6) that vector AO is vector OD, and vector CO is ов; .. line AOOD, CO=0B. Ex. 3. The sides about the equal angles of equiangular triangles are proportionals. Let the triangles ABC, ADE have a common angle A, then, because the angles D and B are equal, DE is parallel to BC. Let vector AD be represented by a, DE by B, then (Art. 3) AB is ma, BC nẞ. .. as vectors, AE = AD+DE= a + ß, AC AB+BC = ma+nß. = Now AC is a multiple of AE, call it p(a+ß). A D E B 2 .. ma+nẞ=p (a + ß), and m=p=n (Art. 6). But line AB: AD=m, line BC: DE =n, .. AB AD :: BC: DE. Ex. 4. The bisectors of the sides of a triangle meet in a point which trisects each of them. Let the sides of the triangle ABC be bisected in D, E, F; and let AD, BE meet in G. Let vector BD or DC be a, CE or EA ß, then, as vectors, BA=BC+CA=2a+2ẞ=2(a + B), DE DC+CE=a+ẞ, = B A F E Now vector BG is along GE, and vector GA along DG. hence CG is in the same straight line with GF, and equal to 2GF. Ex. 5. When, instead of D and E being the middle points of the sides, they are any points whatever in those sides, it is required to find G and the point in which CG produced meets AB. CA m, =n; also let vector DC = a, vector CE=ß; CE Let then · x (ma + ẞ) + y (a + nß). But BA= .. (Art. 6) and xm+y=m, x + yn = n, BG (m − 1) n = BE mn 1 BF=pBA=p (ma + nß). BFBC+CF =ma + a multiple of CG = ma+z{BG – BC'} (ma + B) — ma} . The two values of BF being equated, and Art. 6 applied, Ex. 6. When, instead of as in Ex. 4, where D, E, F are points taken within BC, CA, AB at distances equal to half those lines respectively, they are points taken in BC, CA, AB produced, at the same distances respectively from C, A, and B; to find the intersections. Let the points of intersection be respectively G,, G2, Gз. Ex. 7. The middle points of the lines which join the points of bisection of the opposite sides of a quadrilateral coincide, whether the four sides of the quadrilateral be in the same plane or not. Let ABCD be a quadrilateral; E, H, G, F the middle points of AB, BC, CD, DA; X the middle point of EG. Let vector AB = a, AC = ß, AD =ɣ, then AE + EG= AD + DG gives which being symmetrical is a, ß, y in the same as the vector to the middle point of HF. X is called (Art. 14) the mean point of ABCD. Ex. 8. The point of bisection of the line which joins the middle points of the diagonals of a quadrilateral (plane or not) is the mean |