By Melvin Hausner

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To prove this, first note that if P is constructed so that DCEP is a parallelogram, the mid-point of PC is equal to the mid-point of DE. Hence, but, by construction, Adding (6) and (7) and using (5), we obtain Solving for P, Since a + b + c = 1, c = 1 – a – b, and finally, This technique relates the Cartesian coordinates (x, y) of a point P with its barycentric coordinates. * Choose A as the point (1, 0) on the x-axis, B as (0, 1) on the y-axis, and C = origin = (0, 0) (see Fig. 54). Then by comparing Fig.

D:If P, Q, P′ are given points, there is one, and only one, point Q′ such that . To concentrate more on vectors, rather than on a pair of points which determine the vector, we use a single letter to designate a vector. , exclusively for vectors. Thus from here on we shall often write . The vector will be designated by 0 (read: zero, or the zero vector, depending on how fussy you are). Thus is a way of writing A = B. In handwriting, we usually write for u, for 0, etc. Whenever we identify objects with one another, thereby giving a “definition of equality,” it is crucial to verify three basic “laws of equality”: the reflexive, symmetric, and transitive laws.

The method is to take Q as the center of mass of a suitable physical system. Give A mass 1 and B mass 5. Then P is the location of the center of mass of 1A and 5B (see Fig. 9). By giving C mass 4 we obtain Q as the center of mass of 6P and 4C. The previous examples point the way. By observation, AR/RC = 4/1 and BQ/QR = 5/5 = 1. 9 EXAMPLE 4 (Ceva’s Theorem). Let P be inside ABC. Suppose AP is extended until it meets BC at A1, and that B1 and C1 are found similarly. Then (see Fig. 10), AB1 · CA1 · BC1 = B1C · A1B · C1A The method of “proof” is clear from Fig.