When Is a Geometric Problem
Worth One Hundred Geometric Problems?

Najia Bao

Problem solving is
an important component of learning mathematics, and that topic continues to
receive significant attention in recommendations for school mathematics (NCTM
2000). We should give students much more opportunities to solve geometric
problems with difficult level. Through dealing with a synthetic and difficult
problem, students may review lots of knowledge and methods. The fourteenth
problem in the ninth assignment (see Jim WilsonÕs EMAT 6680) is such a typical
sample. To solve this problem, we need to apply many theorems regarding
midsegment of triangles, isosceles trapezoid, congruent triangles, angle
relationships in circles, four-points-on-a-circle, and so on. The problem is
very helpful for us to develop the divergence, broadness, and flexibility of
our studentsÕ thinking.

Problem 14.

*Construct the Simson line
of a point P (on the circumcircle) and construct the segment connecting P to
the Orthocentner. How do the two segments intersect?*

It is very difficult to solve
this problem about Simson line. So we need to explore by using GSP. As we move
the point P along the circumcircle of the triangle ABC, we notice that the
segment HP is always bisected by the Simson line IJK, that is to say, PX = XH
(see fig. 1).

Figure 1.

Now letÕs analyze this
difficult problem first. In order to prove PX = XH, passing through point H, we
draw a line parallel to segment line IK to construct a triangle PHM. If MJ =
JP, then we get PX = XH (the converse of Midsegment theorem). But to prove MJ =
JP, we must prove quadrilateral HMPQ is an isosceles trapezoid, and then we
need to prove HD = DQ (see
fig. 2).

Figure 2.

Next weÕll start to our proof
of this problem. First, prove HD = DQ. Second, prove that HMPQ is an isosceles
trapezoid. Third, prove MH//IK.

First, prove HD = DQ. Construct auxiliary lines: connect
the points Q and C. If D HDC and D QDC are congruent, then we can get the result that
HD = QD. How to prove that D HDC and DQDC are congruent? We notice that
both of them are right triangles because the point D is the foot of the
altitude AD. Since CF ^ AB and AD ^ BC, the four points F, H, D, and B
are on the same circle. Hence CHD = FBD. And since AQC =ABC, we get CHD = CQD. So D HDC and DQDC are congruent. Therefore, we get
HD = QD (see fig. 3).

Figure
3.

Second, prove that HMPQ is
an isosceles trapezoid. Extend segment PJ to its intersection with the
circumcircle of DABC at point Y. Since YP ^ BC, AQ ^ BC, so YP//AQ. Now we get that the
quadrilateral AYPQ is an isosceles trapezoid, and AY = QP. Passing through
point H, let HM be parallel to AY, and we get HM = AY. So HM = QP and HMPQ is
an isosceles trapezoid. Since JD perpendicularly bisects HQ, JD also
perpendicularly bisects MP, that is, MJ = JP (see fig. 4).

Third, prove MH//IK. Given mBIP = mBJP = 90¡, then the four points I, B, P, and J lie on a
circle with BP as its diameter. So we get that YJI = IBP. And since AYP = ABP, we get that AYP = YJI and AY // IK. Furthermore, we get that MH // IK. Since MJ = JP, then we get that the
segment PH is bisected by IK (see fig. 4).

Figure 4.

To solve the above difficult problem, we
use Òauxiliary circleÓ strategy which is a very important geometric strategy.
By constructing one or more auxiliary circles, we apply the properties of
circle to solve geometric problems. This strategy is often used in finding the
measurement of angles and length of segments, proving equality about angles and
segments, and complicated relationship between segments. It is very useful to
solve considerably difficult problems.