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ABCD EFGH

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Originally appearing in Volume V22, Page 431 of the 1911 Encyclopedia Britannica.
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ABCD EFGH A'B'C'D' = E'F'G'H'. This proves the theorem for parallelograms and also for their halves, that is, for any triangles. As polygons can be divided into triangles the truth of the theorem follows at once for them, 2nd is extended (by the method of exhaustion) to areas bounded by curves by inscribing polygons in, and circumscribing polygons about, the curves. Just as (G. § 8) a segment of a See also:line is given a sense, so a sense may be given to an See also:area. This is done as follows. If we go See also:round the boundary of an area, the latter is either to the right or to the See also:left. If we turn round and go in the opposite sense, then the area will be to the left if it was first to the right, and See also:vice versa. If we give the boundary a definite sense, and go round in this sense, then the area is said to be either of the one or of the other sense according as the area is to the right or to the left. The area is generally said to be See also:positive if it is to the left. The sense of the boundary is indicated either by an arrowhead or by the See also:order of the letters which denote points in the boundary. Thus, if A, B, C be the vertices of a triangle, then See also:ABC shall denote the area in magnitude and sense, the sense being fixed by going round the triangle in the order from A to B to C.

It will then be seen that ABC and ACB denote the same area but with opposite sense, and generally ABC = BCA = See also:

CAB = — ACB = — BAC = — CBA; that is, an inter-See also:change of two letters changes the sense. Also, if A and A' are two points on opposite sides of, and equidistant from, the line BC, then ABC = —A'BC.

End of Article: ABCD EFGH

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