1. Orthocenter

The altitudes {AA',BB',CC'} of triangle ABC pass through a common point, the orthocenter H of the triangle.

The easiest perhaps proof of the concurrency of the three altitudes is to identify them with the medial lines of the anticomplementary triangle A''B''C'', created by drawing parallels to opposite sides. H is then identical with the circumcenter of triangle A''B''C''.

2. Basic properties

[1] Extend CC' to meet the circle again at E. Triangles AHB and AEB are equal since they have common the side AB and equal the angles at the extremities of this side.
[2] Consequently E is symmetric (reflected) to H with respect to side AB.
[3] Extend AO (O circumcenter) to find J on the circumcircle. CHBJ is a parallelogram since its opposite sides are respectively orthogonal to AC and AB.
[4] OK (K middle of AB) has half the length of BJ = CH since it joins the middles of sides of ABJ.
[5] CO and HK meet on at the diametral point L of C since OK is parallel and half of HC.
[6] Let M be the reflexion of O to AB. OM = JB = CH is the vector-sum: OM = OA + OB.
[7] OH is the vector-sum : OH = OM + OC = OA + OB + OC .

3. Exercise

Take 6 points on a circle: N, O, P, Q, R, S. Select three out of the 6. For instance N,O,P and build the orthocenter T of the triangle (NOP). Build the centroid U of the remaining three points. The line [UT] contains the point W, defined through the vector sum: VW = (1/4)(VN+VO+VP+VQ+VR+VS). Thus for all choices of 3 out of the 6 points they result (20) corresponding lines [UT] which pass all through W.

Euler.html
Autopolar.html
Bisector.html
CircleBundleLocus.html
Droz_Farny.html
EulerCircle.html
EulerCircleProperty.html
GaussBodenmiller.html
Isogonal_Conjugation.html
MedialParabola.html
Miquel_Point.html
Orthocenter2.html
OrthoRectangular.html
Polar4.html
RectHypeCircumscribed.html
RectHypeImpossible.html
RectHypeThroughFourPts.html
Simson.html
SimsonDiametral.html
SteinerLine.html
TrianglesCircumscribingParabolas.html
WallaceSimson.html