Download The Nine Point Circle

Robert Weckstein
Lorenzo Grompone
The Nine Point Circle
Statement:
In any triangle, the feet of the altitudes, the midpoints of the sides, and the midpoints of
the line segments connecting the vertices with the orthocenter all lie on a circle.
A
O(a)
H(b)
H(c)
H
M(b)
M(c)
O(9)
O(b)
B
O(c)
H(a)
M(a)
The Nine Point Circle
In order to prove the existence of such a circle, we break the proof into three steps.
C
Lemma 1:
In triangle ABC, points AH and BH are on BC and AC, respectively, so that AAH is
perpendicular to BC and BBH is perpendicular to AC. Prove that triangle AHBHC is
similar to triangle ABC.
B
First notice that the triangles CAB and CBHAH share angle C.
Then, it is clear that angles ABHB and AAHB are equal, as they are both right angles.
Thus quadrilateral ABHAHB is cyclic, and therefore angle ABC supplements angle
ABHAH.
Note that angle ABHAH supplements CBHAH as well.
Thus, angles CBHAH and ABC are congruent, and therefore triangle AHBHC is similar to
triangle ABC.
Lemma 2:
B
In triangle ABC, points AH, BH and CH lie on lines BC, AC, and AB respectively, so that
AAH is perpendicular to BC, BBH is perpendicular to AC, and CCH is perpendicular to
AB. Additionally, point M lies on AB such that segment AM is congruent to BM. Prove
that quadrilateral AHBHCHM is cyclic.
B
By applying the previous proof, we find that angles CBHAH, CBA, and ABHCH are
congruent. We will call this angle value β.
It is obvious that angles CBHAH, ABHCH, and CHBHAH are supplementary and that angle
CHBHAH is equal to 180°-2β.
Additionally, in triangle AAHB, M is the midpoint of the hypotenuse AB. Thus follows
that lines AHM and MB are congruent.
Since lines AHM and MB are congruent, angle MBAH and MAHB are congruent, equal to
angle value β.
Using that fact, BAMAH is equivalent to 180°-2β, and, from this, angle BAMAH is equal
to 2β.
B
B
From this, we can conclude that the quadrilateral MCHBHAH is cyclic, and therefore, the
feet of the altitudes perpendiculars and the midpoints lie on the same circle.
Lemma 3:
B
In triangle ABC, points Ha, Hb, and Hc lie on lines BC, AC, and AB respectively such that
AHa is perpendicular to BC, BHB is perpendicular to AC, and CHC is perpendicular to
AB. Additionally, point H lies on the intersection of lines AHa, BHB, and CHC. Finally,
point Oc lies on line HC such that line HOc is congruent to line OcC. Prove that points Hc,
Hb, Ha, and Oc are concyclic.
B
B
∠ HbHcHa = 180° – 2C Æ Oc lies on the same circle as the midpoints and feet of the
altitudes of triangle ABC because HaHcHbOc is cyclic.
This argument can be repeated for Oa and Ob.
Also, Oa, Ob, and Oc are sometimes referred to as the midpoints of the segments from the
orthocenter to the vertices, and they are:
∠ HbHHa = 360° – C – 90° – 90° = 180° – C Æ HbHHaC is cyclic.
Oc is the circumcenter of HaHbC, and thus the center of the circle circumscribed around
HbHHaC. So COc and OcH are both radii and thus equal in length, and Oc is the midpoint
of HC. Identical proofs for Oa and Ob follow.
Therefore, all nine points lie on the same circle.