Orthocenters and Circumcenters

By Colleen Garrett

I will begin this
investigation by first doing the following:

1. Construct any
triangle ABC.

2. Construct the
Orthocenter H of triangle ABC.

3. Construct the
Orthocenter of triangle HBC.

4. Construct the
Orthocenter of triangle HAB.

5. Construct the
Orthocenter of triangle HAC.

When constructing
the orthocenters of the small triangles we get that the orthocenter of DHCB=A,
orthocenter of DHAB=C, and the orthocenter of DHAC=B.

Now letŐs
construct the circumcircles of triangles ABC, HBC, HAB, and HAC.

CC1, CC2, and CC3
are circumcenters of the small triangles.

LetŐs hide the
circumcircles and construct segments connecting the circumcenters of the small
triangles. Let CC1=D, CC2=E,
CC3=F, and CC4=J. We have a new
triangle DFDE.
Constructing the circumcircle of DFDE we see that it has circumcenter H.
(Recall: H is the orthocenter of
our original DABC).

Hide the
circumcirlce of DDFE.

Claim: DFDE@DABC.

Proof:

Connect
circumcenter F to points C,H, and B. Connect
circumcenter D to points C, H, and A.
Connect circumcenter E to points A,H, and B.

, ,
and because we know that the circumcenter of
a triangle is equidistant from the vertices of the triangle.

We also know that H is the
circumcenter of DDFE so .

Therefore, by transitivity . The result we will be using is .

Now construct a
line perpendicular to through point F. Since ,
F is equidistant from point C and point B; therefore, F must lie on the
perpendicular bisector of . Hence, if
perpendicular to as well. Two lines that are perpendicular to the same line are
parallel; therefore, .

Now construct . We know have two triangles, DDEB and DDCB.

Recall that . is a transversal
intersecting parallel lines and therefore ÐCDBÐEBD by alternating interior angles. Also . We now have DDEBDDCB by SAS.

Now by CPCTC . By a similar argument and . Hence DFDE@DABC by SSS.