The Art and Craft of Problem Solving

8.4 THE POWE:R OF ELEMENTARY GEOMETRY 289

Similar Triangles and Collinear Points

You have noticed, no doubt, that similar triangles are crucial ingredients in almost ev­
ery geometry problem. Finding the "crux" similar triangles can be tricky. Sometimes
you have to conjure up similar triangles out of thin air, and sometimes you have to
wade through a complex and cluttered diagram to find the "natural" correspondences
that shed genuine light on the problem. Here are some examples.

Example 8.4.4 Let ABC be an isosceles triangle with AB = AC. Drop a perpendicular
from A to BC, meeting BC at D. Then drop a perpendicular from D to AC, meeting it

at E. Let F be the midpoint of ED. Prove that AF ..1 BE.

A

B c

Solution: Angle chasing is tempting, as always, but it is doomed to failure here.
One of the intersecting lines of interest, AF, is a median of triangle AED. Medians, in
contrast, say, to angle bisectors or altitudes, do not yield useful angular information,
without very messy trigonometry.
We need to think strategically. What is a penultimate step for perpendicular lines?

Inscribed right triangles! By 8.2.17, if a side of a triangle is a diameter of a circle, and

the opposite vertex is on the circle, then that vertex is a right angle. Notice that ADB
is such an animal: a right triangle inscribed in the circle with diameter AB. Thus, we
will be done if we can prove that G lies on this circle, for then AGB would also be a
right triangle.

In other words, we must show that A, G, D, Bare concyclic points. By 8.2.18, this

is true if and only if

LGAD=LGBD.

At this point, almost tasting success, we'd be silly not to try more desperate angle
chasing. But we need more: we have to incorporate the fact that F is the midpoint of
DE in an "angular" way. We need another triangle that interacts with a midpoint in a
similar (pun intended) fashion.
Notice that AF is the median of a right triangle. Are there any other midpoints
floating around? Yes, D is the midpoint of BC, since ABC is isosceles-the first time
we used this fact! If only DE were a midline of a triangle. Wishful thinking comes to
the rescue: Draw line BH parallel to DE.