8.4 THE POWER OF ELEMENTARY GEOMETRY 285AB' = AB cosA. Substituting, we have
AM^2 =AC'·AB = AC·ABcosA
and
AP^2 =AB'· AC =AB ·ACcosA,
and we're done! •But we promised two solutions! Let's look at option #4, using the converse of
POP. At first, it seems complicated, because POP requires four different segments.However, we have two circles to work with, so we have a decent chance of using POP
several times on overlapping circles. We need to label the now-critical intersection
of QP and MN ; call it X in the diagram below. Notice that X is the orthocenter of
ABC. Also observe-thanks again to the carefully drawn diagram-that COl, IDl, and
CB intersect at a single point, which we naturally label A', since it is none other than
the foot of the altitude through A. (The three altitudes meet at X, and AA' B is a right
triangle inscribed in IDl, while AA'C is a right triangle inscribed in COl.)
We need to prove that QX. X P = MX. X N. These four segments do not live in a
single circle, unfortunately. However, the dashed line AXA' saves the day, since it is a
chord in both COl and IDl.Applying POP to COl, we have
AX · XA' = QX · XP.
But in IDl, POP yields
AX · XA' =MX ·XN,
and once again, we're done. •Notice how much easier the POP solution was. But the first solution was not too
bad, either. Both required some bravery, careful diagrams, and systematic observations
about perpendiculars, right triangles inscribed in circles, etc. Also, both solutions
hinged upon finding the "crux object." In the first solution, we zeroed in on point A. In