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#6

### How do I know this Information?

Filling in a Few "Gaps"

**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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#6.*

**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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#6.*