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.

(i) The perpendicular bisectors of the three sides of a triangle meet at
a point, called the

circumcenter of the triangle. (ii) There is a unique circle that
passes through the vertices of a triangle, and the center of this circle

(the circumcircleof the triangle) is the circumcenter.

Proof.

Let the triangle be ABC, as shown, and let M, N be the midpoints of BC andAC, respectively. Also let the perpendicular bisectors of BC, AC be m and n,

respectively. Let

m and n meet at O.
Since

O lies on the perpendicular bisector of BC, OB = OC (Perpendicular Bisector Theorem). Similarly,OC = OA. Together, we have OB = OA = OC, i.e.,

O is equidistant from A, B and C. In particular, O is equidistant from A and .B

By the Converse of the Perpendicular Bisector Theorem, O lies on the perpendicular bisector of AB. As

O 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

O that passes through the vertices A, B, and C, and that any such circle must
coincide with this circle. Consider the circle

K with center O and radius OA. K
must pass through

B 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

K0 passing throughA, B and C must coincide with K. Let the center of K0 be O0. Since O0 is

by definition equidistant from

B and C, O0 lies on the perpendicular bisector of BC(Converse to PBT), and therefore lies on
must also lie on

n, the perpendicular bisector of AC. Therefore O0 is the point of
intersection of

m and n, which is O. This shows O = O0. Since K0 passes through A,
the radius of

K0 is also OA. Hence K0 = K because 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 ~~BC~~ and ~~BC~~ and ~~AC~~ are perpendicular to ~~BC~~ | | ~~AC~~. But

*m*and*n*don't intersect -- that is, that they are parallel. It's given that*m*is perpendicular to*n*is perpendicular to*AC*. So, by the Fifth Postulate, both*m*, so, by the Two Perpendiculars Theorem,*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!

The three altitudes of a triangle meet at a point, called the orthocenter

of the triangle.

Proof.

Let the altitudes of triangle ABC be AD, BE 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

AC0BC, ABCB0 are parallelograms. Therefore,

by the Parallelogram Theorem (opposite sides are congruent)

C0A = BC = AB0. Thus A 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 C0B0. Similarly,

Line FC and Line BE are perpendicular bisectors of A0B0 and C0A0, respectively.

By Theorem 13,

Line AD, Line 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

.

## No comments:

## Post a Comment