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 line is given a sense, so a sense may be given to an area. This is done as follows. If we go round the boundary of an area, the latter is either to the right or to the 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 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 positive if it is to the left. The sense of the boundary is indicated either by an arrowhead or by the order of the letters which denote points in the boundary. Thus, if A, B, C be the vertices of a triangle, then 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 = CAB = — ACB = — BAC = — CBA; that is, an interchange 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.
Taking account of the sense, we may make the following statement
If A, A' are two corresponding points, if the line AA' cuts the axis in B, and if C is any other point in the axis, then the triangles ABC and A'BC are corresponding, and
End of Article: ABCD EFGH 

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