The Art and Craft of Problem Solving

278 CHAPTER 8 GEOMETRY FOR AMERICANS

triangles CAF and DAG are similar. (If two right triangles have a single pair of equal

angles, they will be similar, since already the two right angles are equal!)

But CAF r-v DAG implies that u/v = b/m. Hence

Therefore,

x [ACD] b

Y [ADB] c

The converse of this theorem is also true; i.e., if D is placed on BC so that

CD/DB = AC/AB, then AD bisects LCAB. We leave the proof of this important fact

to you (Problem 8.3.28).

Example 8.3.13 Proof of the Centroid Theorem (8. (^1) .3). Let us start by temporarily
removing one of the medians (after all, we do not know if the three medians actually

meet in a single point) and adding an auxiliary line connecting the midpoints E and F.

A

c B

Line EF is sometimes called a midline of triangle ABC, and cuts off triangle AEF

which is clearly similar to ABC by 8.3.9. The ratio of similitude is 1 : 2, so EF : AC =

1 : 2.

Furthermore (also by 8.3.9), EF is parallel to AB. Thus LFEG = LGCA and

LEFG = LCAG, so D"FEG r-v D"GCA. Again, the ratio of similitude is 1 : 2, since

EF : AC = 1 : 2. Consequently, EG : GC = FG : GA = 1 : 2.

There was nothing special about the two medians that we used; we have shown

that the intersection of any two medians cuts them into segments with a 1 : 2 ratio.

When we draw the third median BD, it will intersect median AF in a point, say G',

such that G'D: G'B = G'F : G'A = 1 : 2. But there is only one point that divides me­

dian AF in this way, namely G. So G' = G and we are done. _

Note the interplay of ratios, parallels, and similar triangles, and the subtle use of a

"phantom point" (G') to show that the intersection point was unique.