The Art and Craft of Problem Solving

A

8.3 SURVIVAL GEOMETRY II 277

B

This gives us two similar triangles: ACE rov ABD. What do we do next? Investigate the
diagram further, and hope that there is more information. We now have ratios to play
with.

Remember, we want to prove that CD/DB = AC/AB. Using our similar triangles,

we know that AC/AB = CE/DB. If we can prove thatCE = CD, we'll be done!

How to show CE = CD? We need to show that triangle CED is isosceles. This

follows from easy angle chasing. It is easy to check that L.CDE = a + /3, since it

is an exterior angle of triangle ABD. Likewise L.CED = a + /3. So triangle CED is

isosceles, and we're done. _

The second proof also uses similar triangles, but the crux idea involves area. We

want to investigate the ratio of lengths CD /DB =x/y. (See the figure below for labeled

lengths.) We equate this with a ratio of areas, by using 8.3.4. Triangles ACD and ADB

share the vertex A, so the ratio of their areas is

[ACD] x

[ADB] y

Next, we compute the two areas a different way, so as to involve the lengths AC = b

and AB = c. Let rn = AD be the length of the angle bisector. Consider the altitudes

u = CF and v = DG as shown. Then

[ACD]

[ADB]

1 2 urn

lvc 2

urn

vc

B

Now let's get a handle on the ratio u/v. This suggests looking at similar triangles,

and indeed, because AD is the angle bisector, L.CAF = L.DAG, and thus the two right