Why is the area of triangle XZW one-third the area of triangle XYZ, and why is the area of triangle FDC one-fourth the area of triangle ABC?
In any triangle, a median splits the triangle into two triangles of equal area. For example, consider triangle ABC with median CF. The small triangles, AFC and CFB, have the congruent bases (AF and FB.) But they also have congruent heights (the distance between C and AB, or the distance between the line parallel to AB drawn through point C.) Therefore they must have the same area.
Then consider triangle AFC with median FD. For the same reasons, triangles AFD and FDC have the same area, and so the area of triangle FDC is one-fourth the area of triangle ABC.
Similar reasoning can be used to see that the area of triangle XZW is one-third the area of triangle XYZ: since W and Q trisect XY, the bases of triangles XZW, WZQ, and QZY are all congruent. These triangles also have congruent heights (the distance between the line parallel to XY through Z. Since the three triangles have equal area, the area of XZW is one-third the area of XYZ.
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Why does SW = ZV = (1/3)BD and FH = GD = (1/3)BD? (And so SW = FH.) Here's one way to see it:
Since SW is a segment joining a pair of trisection points on XZ and XY, it must be parallel to and equal one-third of ZY. Of course, ZV = (1/3)ZY = (1/3)BD, since ZY was constructed to be the same length as median BD.
In triangle ABC, notice that H is the midpoint of AJ, so segment FH joins the midpoints of two sides of triangle AJB. Therefore, FH is parallel to BD. Consequently, FHDG is a parallelogram, and FH = GD = (1/3)BD. Thus, SW = FH.
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