Difference between revisions of "Orthocenter"
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That seems somewhat overkill to prove the existence of the orthocenter. We use a much easier (and funnier) way. | That seems somewhat overkill to prove the existence of the orthocenter. We use a much easier (and funnier) way. | ||
− | Let the line through <math>B</math> parallel to <math>AC</math> and the line through <math>C</math> parallel to <math>AB</math> intersect at <math>D.</math> Define <math>E,F</math> similarly. Note that <math>FA=BC=AE,</math> so the <math>A</math> altitude of <math>\triangle ABC</math> is the perpendicular bisector of <math>DE.</math> Since the circumcenter exists, the | + | Let the line through <math>B</math> parallel to <math>AC</math> and the line through <math>C</math> parallel to <math>AB</math> intersect at <math>D.</math> Define <math>E,F</math> similarly. Note that <math>FA=BC=AE,</math> so the <math>A</math> altitude of <math>\triangle ABC</math> is the perpendicular bisector of <math>DE.</math> Since the circumcenter exists, the orthocenter must too. |
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Revision as of 10:25, 31 December 2021
The orthocenter of a triangle is the point of intersection of its altitudes. It is conventionally denoted .
The lines highlighted are the altitudes of the triangle, they meet at the orthocenter.
Proof of Existence
Note: The orthocenter's existence is a trivial consequence of the trigonometric version Ceva's Theorem; however, the following proof, due to Leonhard Euler, is much more clever, illuminating and insightful.
Consider a triangle
with circumcenter
and centroid
. Let
be the midpoint of
. Let
be the point such that
is between
and
and
. Then the triangles
,
are similar by side-angle-side similarity. It follows that
is parallel to
and is therefore perpendicular to
; i.e., it is the altitude from
. Similarly,
,
, are the altitudes from
,
. Hence all the altitudes pass through
. Q.E.D.
This proof also gives us the result that the orthocenter, centroid, and circumcenter are collinear, in that order, and in the proportions described above. The line containing these three points is known as the Euler line of the triangle, and also contains the triangle's de Longchamps point and nine-point center.
Easier proof
That seems somewhat overkill to prove the existence of the orthocenter. We use a much easier (and funnier) way.
Let the line through parallel to
and the line through
parallel to
intersect at
Define
similarly. Note that
so the
altitude of
is the perpendicular bisector of
Since the circumcenter exists, the orthocenter must too.
Properties
- The orthocenter and the circumcenter of a triangle are isogonal conjugates.
- If the orthocenter's triangle is acute, then the orthocenter is in the triangle; if the triangle is right, then it is on the vertex opposite the hypotenuse; and if it is obtuse, then the orthocenter is outside the triangle.
- Let
be a triangle and
its orthocenter. Then the reflections of
over
,
, and
are on the circumcircle of
:
- Even more interesting is the fact that if you take any point
on the circumcircle and let
to be the midpoint of
, then
is on the nine-point circle.