The Art and Craft of Problem Solving

290 CHAPTER 8 GEOMETRY FOR AMERICANS

B c

Now BHC is a right triangle with midline DE and median BE. And sure enough, it is

obvious that

l:::.BHC"" l:::.AED.

It immediately follows (using the similar subtriangle reasoning of Problem 8.3.30) that

l:::.FAD"" l:::.EBC,

so LGAD = LGBD, as we hoped. •

Drawing the auxiliary line BH was a clever, albeit very natural crux move. In the

next example, we have much more clutter in our diagram, with many more choices, so

we must take special care to think about the "natural" auxiliary object, which in this

case is not a line, but a triangle.

Example 8.4.5 The Euler Line. Prove that the orthocenter, circumcenter, and centroid

of any triangle are collinear, and that moreover, the centroid divides the distance from

orthocenter to circumcenter in the ratio 2 : I. These three points determine a line called

the Euler line of the triangle.

Solution: How to prove this? It is important to draw a careful diagram, as always,

and also, we must try not to clutter up the diagram. Here is a start. The orthocenter,

centroid, and circumcenter of triangle ABC are respectively, H, M, and K. We drew in

two altitudes to fix H, but we did not draw any median lines, circumradii, or perpendic­

ular bisectors. We did include one auxiliary construction: we connected the midpoints

E, G, and I of the sides of triangle ABC to form the medial triangle EGI. Indeed, this

is our crux move. If you don't know why, you probably have not done Problem 8.2.2 9

yet. So do it now, before you read further!