Friday, November 14, 2014

The Concurrency Theorems (Day 64)

I wanted to give these concurrency theorems during Chapter 5. But that unexpected week off trying to align my blog's calendar to my district's calendar ruined my Chapter 5 plans. So I put all of the concurrency theorems right here, since I can afford to put them here without ruining Chapter 6.

As it turns out, Hung-Hsi Wu proves all four of these concurrency theorems, as opposed to the U of Chicago, which only discusses one of them. We know enough to prove two of them using results from Chapter 5 and before -- the circumcenter and orthocenter -- the other two are more difficult and must wait until Chapter 7. Therefore, most of this post will be cut-and-paste from the proofs given by Wu.

We begin with the circumcenter, the intersection of the perpendicular bisectors. This is the only concurrency theorem mentioned in the U of Chicago. (The theorem numbering is Wu's.) The first part of Wu's proof is nearly identical to the U of Chicago's:

Theorem 13.
(iThe perpendicular bisectors of the three sides of a triangle meet at
a point, called the
circumcenter of the triangle. (iiThere is a unique circle that
passes through the vertices of a triangle, and the center of this circle
(the circumcircle
of the triangleis the circumcenter.
Let the triangle be ABC, as shown, and let Mbe the midpoints of BC and
AC, respectively. Also let the perpendicular bisectors of BCAC be m and n,
respectively. Let
m and n meet at O.
lies on the perpendicular bisector of BCOB OC (Perpendicular Bisector Theorem
). Similarly,
OC OA. Together, we have OB OA OC
is equidistant from Aand C. In particular, is equidistant from and .
By the Converse of the Perpendicular Bisector Theorem, lies on the perpendicular bisector of AB
already lies on the perpendicular bisectors of BC and AC, this proves the rst
part of the theorem.
To prove the second part, we have to rst prove that there is a circle with center at
that passes through the vertices AB, and C, and that any such circle must
coincide with this circle. Consider the circle
with center and radius OAK
must pass through
and C, on account of OB OA OC, so we have proved
the existence of such a circle. Next,we have to show that any other circle
Aand must coincide with K. Let the center of Kbe O0. Since Ois
by definition equidistant from
and COlies on the perpendicular bisector of BC
(Converse to PBT), and therefore lies on
m. For exactly the same reason, O0
must also lie on
n, the perpendicular bisector of AC. Therefore Ois the point of
intersection of
m and n, which is O. This shows O0. Since Kpasses through A,
the radius of
Kis also OA. Hence Kbecause they have the same center and
the same radius. The proof of the theorem is complete.

But then, as I pointed out earlier, part of this proof depends on some Parallel Postulate. I want, in fact, to use our version of the Fifth Postulate, since all that's needed for the proof was the part about Perpendicular to Parallels. But unfortunately, I couldn't find a way to avoid an indirect proof. (Wu, in a Remark, also refers to this fact that m and n don't necessarily intersect at any point, much less O, without a Parallel Postulate. In non-Euclidean hyperbolic geometry, this theorem is in fact false.)

Assume that m and n don't intersect -- that is, that they are parallel. It's given that m is perpendicular to BC and n is perpendicular to AC. So, by the Fifth Postulate, both BC and AC are perpendicular to m, so, by the Two Perpendiculars Theorem, BC | | AC. But C is on each line. So lines BC and AC are identical, and the three points are collinear. But A, B, and C are the vertices of a triangle, so they can't be collinear. This is a contradiction. Therefore m and n intersect. QED

Notice the similarity between this and the proof of the Two Reflection Theorem for Translations. I point out that the key difference is that in the latter, we were trying to prove that three points are collinear, whereas here, we use the fact that three points aren't collinear to derive a contradiction in an indirect proof.

Now for the other proof -- the orthocenter proof. I will cut and paste Wu's proof. In his proof, he ends up showing that the orthocenter of a triangle is the circumcenter of a larger triangle!

Theorem 14.
The three altitudes of a triangle meet at a point, called the orthocenter
of the triangle.
Let the altitudes of triangle ABC be ADBE and CF. Through each vertex,
draw a line parallel to the opposite side, resulting in a triangle which we denote by
triangle A0B0C0
By construction, the quadrilaterals
AC0BCABCBare parallelograms. Therefore,
by the Parallelogram Theorem (opposite sides are congruent)
C0A BC AB0. Thus is the midpoint of
C0B0. Moreover, since AD perpendicular BC and BC | | C0B0, we also know that AD perpendicular C0B0
(by our Fifth Postulate). It follows that
Line AD is the perpendicular bisector of C0B0Similarly,
Line FC and Line BE are perpendicular bisectors of A0Band C0A0, respectively.
By Theorem 13,
Line ADLine FC and Line BE meet at the circumcenter of triangle A0B0C0. The proof
is complete.

I provide no exercises for this lesson/activity -- since the proofs are difficult enough on their own


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