The Art and Craft of Problem Solving

276 CHAPTER 8 GEOMETRY FOR AMERICANS

We need to include two obvious auxiliary lines (XY' and X'Y) in order to find similar

triangles.

Indeed, these similar triangles are very easy to find. Whenever two triangles share an

angle, then they will be similar if we can find just a single pair of equal angles (since

that will force the third pair of angles to be equal, since the sum of the angles in each

triangle is 180°).

So our candidates are the two triangles that contain angle P. The next step is to

apply 8.2.16, which says that inscribed angles in a circle are equal to half the arc they

subtend. The measure of the inscribed angles is generally of less importance than the

fact that

Two inscribed angles that subtend the same arc must be equal.

Consequently, angles Y and Y' are equal, because they subtend the same arc XX'. We

conclude that

PYX' rv PY'X.

(Note that the order of the points is crucial. Triangles PX'Y and PXY' are not similar. )

It follows that

PX' /PY = PX/PY',

and cross-multiplying yields

PX ·PY = PX'·Py'.

There are two other configurations, since P may also be on or inside the circle. In

the first case, the products simply equal zero. The second case is handled in exactly

the same way as before, by drawing two obvious auxiliary lines and then finding two

similar triangles. _

Example 8.3. 12 Two Proofs of the Angle Bisector Theorem (8. (^1) .2). There are many
ways to prove this. We will give two proofs with very different flavors.
The first proof uses the strategy of creating similar triangles out of thin air. We
are given an angle bi sector, so we already have two triangles with one pair of equal
corresponding angles. If only we had another pair of equal angles! We are given that

LCAD = LDAB = a. Our trick is to find point E on AD so that LACE = LABC, which

we denote by f3.