The Art and Craft of Problem Solving

8.4 THE POWER OF ELEMENTARY GEOMETRY 287

But there are other area ratios known to us. Since

[CAF]: [CFB] = I: 2,

and

[CFB] = 6y + 2x,

we see that [CAF] = 3y+x, and thus [CIA] = 3y.

Now we can go in the reverse direction. The ratio of areas

[CIA] : [CID] = 3y : 4y = 3 : 4

yields a ratio of lengths

AI: ID = 3: 4.

We're not done. We can extract more information in this way. Since

[BAD] : [CAD] = I : 2,

we have

2(3x+2y) = 3y +4y,

so 2x = y. This means that

[FIA] : [CIA] = 1 : 6,

so we conclude that

FI :IC = 1: 6.

Going back to the original diagram, with three cevians, we see that each is divided

in the ratio 1 : 3 : 3. For example, FI : IH : HC = I : 3 : 3. More important, the

vertices G, H, I of the shaded triangle bisect, respectively, the segments H B, CI, AG.

This situation is shown without any clutter (but with some important auxiliary lines)
below.

The auxiliary lines should make it evident that the large triangle has been dissected

into seven triangles, each with area equal to that of the shaded triangle! So we con­

clude that [GHI] = 1/7. •