What is the formula for orthocenter

What are heights in the triangle?

......If the perpendicular falls from a corner of any triangle to the opposite side, its height is created.

The height describes how high the triangle is.

Since there are three corner points and three sides, there are also three heights.

Calculating the heights Top
There is an angle and a matching side.

......A height, for example the height hc, divides a triangle into two right triangles.
In the right triangle, h appliesc= a * sin (beta), in the left hc= b * sin (alpha).
So there are two ways to change the height hc to calculate.
The following two formulas are derived accordingly,
Ha= b * sin (gamma) = c * sin (beta) and hb= c * sin (alpha) = a * sin (gamma)

Sine law
From hc= a * sin (beta) and hc= b * sin (alpha) follows a * sin (beta) = b * sin (alpha) or a: b = sin (alpha): sin (beta).
The formula is expanded to a: b: c = sin (alpha): sin (beta): sin (gamma).
This is the law of sines, which makes it possible to calculate the fourth out of three pieces of any triangle.
The four pieces are two angles and their opposite sides.

The three sides are given.
...The formulas apply
Ha= (2 / a) sqrt [s (s-a) (s-b) (s-c)],
Hb= (2 / b) sqrt [s (s-a) (s-b) (s-c)],
Hc= (2 / c) sqrt [s (s-a) (s-b) (s-c)].
Where s = (1/2) (a + b + c).
For proof
The formula by Heron A = sqrt [s (s-a) (s-b) (s-c)] applies to the area of ​​a triangle.
Where s = (1/2) (a + b + c).
A proof can be found e.g. at Arndt Brünner (URL below).

If we put A = (1/2) chc and solves the Heron formula according to hc the third formula results.
The other two formulas are derived accordingly.

Intersection Top
Existence of the height intersection

......Three heights meet at exactly one point, the height intersection point H.
The usual evidence from the textbooks follows.

......The triangle is completed by drawing parallels to the sides through the corner points.

Then the given triangle becomes the center triangle of a larger triangle.

......If the heights of the given triangle are lengthened within the large triangle, they become their perpendiculars. The vertical intersection becomes the intersection of the perpendiculars.
However, the three perpendiculars intersect at one point, since AM = BM and BM = CM for two perpendiculars and therefore AM = CM for the third.

The intersection of the height of the obtuse triangle
.........If the triangle is obtuse, the point of intersection of the heights is outside the triangle.
Because of this peculiarity, only the acute-angled triangle is considered on this page, although statements mostly also apply to the obtuse-angled triangle.
The evidence requires a new presentation, which I will spare myself.

In the case of a right triangle, the vertical intersection coincides with the vertex of the right angle.

Excellent points in the triangle Top
In addition to the heights, other lines of the triangle also intersect at a point.

There are connections between them. Some will be shown on this website.

Area formulas Top

......As already mentioned, the area of ​​any triangle can be determined with the help of a height.

We have A = (1/2) chc or A = (1/2) aha and A = (1/2) bhb.
To prove this, the triangle is transformed into a rectangle of the same area.

A new formula results from the area formula.
......The relationship is aha = brab = chc

The product of one side and its height is constant.

The area of ​​a triangle can also be calculated from the radius of the inscribed circle.
......We have A = (1/2) ar + (1/2) br + (1/2) cr.
Then 2A = ar + br + cr or 1 / r = [1 / (2A)] (a + b + c) or 1 / r = a / (2A) + b / (2A) + c / (2A) .
Because A = (1/2) aha, A = (1/2) bhb and A = (1/2) chc is 1 / r = 1 / ha + 1 / hb+ 1 / hc.
That's another formula.

Elevation sections Top
The heights divide the triangle into six sub-triangles, of which three pairs are similar.

The yellow triangle exists six times. This means that there are a total of 6 * 3 = 18 pairs of similar triangles.
This means that there are also numerous formulas between the 6 height and 6 side sections, because with similar triangles, the corresponding sides are in the same ratio.

The following case is of interest.
......An example of similar triangles are the triangles AHHc and CHHa

The proportion CH applies to the marked sides:HHa = AH:HHc.
The associated product equation is CH * HHc = AH * HHa.

If you refer to the third height brab with one, it is CH * HHc = AH * HHa = BH * HHb
The product of the two height sections of a height is constant.

Euler straight line Top

......The Euler line is a straight line of the triangle that runs through the circumcenter M and the center of gravity S.
It is noteworthy that the intersection of the heights H also lies on this straight line.
In addition, HS = 2 * SM applies.

The proof is done in six steps.
Preliminary remark
......If one wants to prove that the three marked points lie on a straight line, it is advantageous for computational reasons to divide the triangle in a Cartesian coordinate system not by the sides a, b and c, but by the height hc= r and that by the height hc Define generated route sections p and q.
Let p> q hold.
This makes the calculations relatively easy.

Height intersection point H
......The point H has the x-coordinate 0.
The y-coordinate h is calculated with the help of the two colored, similar triangles.
The following applies: h: q = p: r or hr = pq or h = (pq) / r.
Result: The point has the representation H (0 | (pq) / r) or H (0 | h).

Center of the circle M
The center point of the circumference is determined by the vertical perpendicular.
......The middle of the page is shown as Ma(p / 2 | r / 2) and Mc[(p-q) / 2 | 0].
The x coordinate of Mc results from (p + q) / 2-q = (p-q) / 2.
The straight line MMa stands vertically on side a and therefore has the slope p / r. She goes through point Mc and therefore has with the approach (y-y1) = m (x-x1) the equation (y-r / 2) = (p / r) (x-p / 2) or y = (p / r) x-p² / 2r + r / 2.
If one sets for x = (p-q) / 2, then M results for the y-coordinate of the center point
y = (p / r) (pq) / 2-p² / (2r) + r / 2 = p² / (2r) -pq / (2r) -p² / (2r) + r / 2 = r / 2- ( pq) / (2r) = r / 2-h / 2 = (rh) / 2

Result: We have M [(p-q) / 2 | (r-h) / 2].

Focus S
The center of the center of gravity is determined by the bisector.
They intersect at a ratio of 2: 1.
......This partial ratio is transferred to the coordinates of the center of gravity after the first line of rays.
The center of gravity S has the y-coordinate r / 3.
The horizontal side of the yellow triangle has the length (p + q) / 2-q = (pq) / 2 in the x-direction and point S the x-coordinate (2/3) (pq) / 2 = (pq) / 3.
Result: We have S [(p-q) / 3 | r / 3].

It remains to be shown that the points H, M and S lie on a straight line.
The straight line equation of the straight lines through H (0 | h) and M [(p-q) / 2 | (r-h) / 2] is determined according to the two-point form.
(y-y1) / (x-x1) = (y2-y1) / (x2-x1)
<=> (y-h) / x = {[(r-h) / 2] -h} / [(p-q) / 2]
<=> (y-h) / x = (r-h-2h) / (p-q)
<=> y ​​= [(r-3h) / (p-q)] x + h
If you insert the x-coordinate of the point S [(p-q) / 3 | r / 3], namely x = (p-q) / 3, y = r / 3 results.
This means that S is also on HM.

A length relationship
What remains is HS = 2 * SM.
The lengths of the lines HM and SH are calculated using the formula s = sqrt [(x2-x1) ² + (y2-y1) ²] calculated.
......HM = sqrt {[(p-q) / 2] ² + [((r-h) / 2-h] ²} = (1/2) sqrt [(p-q) ² + (r-3h) ²]

SH = sqrt {[(p-q) / 3] ² + (r / 3-h) ²} = (1/3) sqrt [(p-q) ² + (r-3h) ²]

From this it follows 2HM = 3SH or 2 * (SH + SM) = 3 * SH or 2 * SM = SH, wzbw.

Feuerbach district Top
In the English-speaking area it is called the nine-point circle.

......The Feuerbach circle runs through 9 points, namely through the height base points H.a, Hb, Hc, the midpoints of the sides Ma, Mb, Mc and the midpoints of the height sections H emanating from the corner points1, H2 and H3
The center F of the Feuerbach circle is the bisection point of HM, where H and M are the intersections of the heights and the perpendicular.

The proof is done in eight steps.
Preliminary remark
......Obviously it suffices to show that the circle passes through the points related to side c.

These are the points Mc , Hc, and H3.

Coordinates of the center of the Feuerbach circle
First of all, it is about the coordinates of the center of the Feuerbach circle F.
They result from the formulas x = x1+ (1/2) (x2-x1) and y = y1+ (1/2) (y2-y1), which for the midpoint of a segment P1P.2 be valid.
With H (0 | h) and M [(pq) / 2 | (rh) / 2] this leads to x = 0 + (1/2) (pq) / 2] = (1/4) (pq) and y = h + (1/2) [(rh) / 2-h] = (1/4) (h + r).
Intermediate result: We have F [(1/4) (p-q) | (1/4) (h + r)]

The radius of the Feuerbach circle is H.C.F and is based on the formula P1P.2== sqrt [(x2-x1) ² + (y2-y1) ²] calculated.
With Hc(0 | 0) is H.cF² = [(1/4) (p-q)] ² + [(1/4) (h + r)] ².

Circular equation
The circle equation is calculated using the formula (x-xm) 2 + (y-ym) ² = r², where the center point M (xm| ym) and the radius are r.
The result is [x- (1/4) (pq)] ² + [y- (1/4) (h + r)] ² = [(1/4) (pq)] ² + [(1/4 ) (h + r)] ².

It is to be shown by inserting that the points Mc , Hc, and H3 lie on the circle.

If one sets x = y = 0, then the same terms result on the left and right. The circle equation is fulfilled.
[0- (1/4) (pq)] 2 + [0- (1/4) (h + r)] 2 = [(1/4) (pq)] 2 + [(1/4) (h + r)] ²

The y-coordinate is y = 0
The x-coordinate is x = p- (p + q) / 2 = p / 2-q / 2 = (p-q) / 2.
The term [x- (1/4) (pq)] in the circular equation is then x- (1/4) (pq) = (1/2) (pq) - (1/4) (pq) = (1 / 4) (pq).
This results in the same terms on the left and right in the circular equation. The circle equation is fulfilled.

The x-coordinate is x = 0.
The y-coordinate is y = h + (r-h) / 2 = h / 2 + r / 2 = (r + h) / 2
The term [y- (1/4) (h + r)] in the circular equation is then (1/2) (h + r) - (1/4) (h + r) = (1/4) (r + h).
This results in the same terms on the left and right in the circular equation. The circle equation is fulfilled.

Appropriate considerations
The same considerations apply to the points referred to on pages a and b.
That’s the proof.

Feuerbach district and area
Above it is shown that for the radius of the Feuerbach circle HcF² = [(1/4) (p-q)] ² + [(1/4) (h + r)] ².
More interesting than this term is the statement that the radius of the perimeter is twice as large.
To prove this, the formula s² = sqrt [(x2-x1) ² + (y2-y1) ²] the radius is determined by the square of the circumference.
With A (0 | 0) and M [(pq) / 2 | (rh) / 2] the equation AM² = [(1/2) (pq)] ² + [(1/2) (h + r) applies )] ².
The assertion follows from this.

Elevation triangle Top

......The triangle made up of the base points is called the height base point triangle.

It has some interesting properties.

Inscribed circle of the height base triangle
......The following applies:
The point of intersection of the heights of a triangle is also the center of the inscribed circle of its height base triangle.

In other words: the heights of any triangle coincide with the bisector of its height base triangle.

......The yellow square AHcHHb is a chordal quadrilateral because the interior angles at the opposite points Hb and Hc are right angles and thus complement each other to 180 °. Further, the angles marked in red are the same, since there are angles over the same chord HHb are.
With the same conclusion, the blue angles are above the chord HHa equal.

......Consider the right triangles AHHb and HBHa
Because of the equal vertex angles, they are similar.
So the blue and red angles are the same size.
So the angle becomes HbHcHa by HHc halved.

In the same way one shows that the other two interior angles at Ha and Hb of the height base triangle are halved by the heights.

That is the proof.
Interior angle
The interior angles of the height base triangle are 180 ° -2 * alpha, 180 ° -2 * beta, and 180 ° -2 * gamma.
......The straight line AHa is the height of the triangle and at the same time the bisector of the angle HbHaHc
If you reflect on it, then the two angles, which are marked with alpha ', are the same. This is anticipated in the drawing.
Correspondingly, the angles beta 'and gamma' are the same.

......The sum of the angles in the three yellow triangles is 180 °. This means:
(I) alpha + beta '+ gamma' = 180 °
(II) alpha '+ beta + gamma' = 180 °
(III) alpha '+ beta' + gamma = 180 °
(I) - (II) gives alpha + beta'-alpha'-beta = 0 or (IV) alpha'-beta '= alpha-beta.
From (III) follows (V) alpha '+ beta' = 180 ° -gamma.
(IV) + (V) gives alpha'-beta '+ alpha' + beta '= alpha-beta + 180 ° -gamma or alpha = alpha'.
Correspondingly, one shows that beta = beta 'and gamma = gamma'.

......So the yellow triangles and the starting triangle are similar.
That is a by-product.
The interior angles of the height base triangle can be read off in the drawing:
180 ° -2 * alpha, 180 ° -2 * beta, and 180 ° -2 * gamma.

The formulas H apply to the sidesaHc= b * cos (beta), HaHb= c * cos (gamma) and HbHc= a * cos (alpha).
It should be the side x = HcHa according to the formula s = sqrt [(x2-x1) ² + (y2-y1) ²].
......For this you need the coordinates of Hc and Ha. It is Hc(0|0).
To calculate the coordinates of Ha bring the straight line AHa and BC to the cut.
BC: y = - (r / p) x + r
AHa: y = (p / r) x + h, where h = (pq) / r
This leads to - (r / p) x + r = (p / r) x + (pq) / r = ... = x = (pr²-p²q) / (p² + r²).
The following applies: y = - (r / p) x + r = - (r / p) (pr²-p²q) / (p² + r²) + r = (pqr + p²r) / (p² + r²).
Then s² = [(pr²-p²q) / (p² + r²)] ² + [(pqr + p²r) / (p² + r²)] ² = ... = [p² / (p² + r²) ²] * [ (q² + r²) (p² + r²)].
With a² = p² + r² and b²q² + r² and p = a * cos (beta), s² = (p² / a4) (a²b²) = (p² / a²) b² or s = (p / a) b.
With p = a * cos (beta), s = b * cos (beta) = HaHc.
Corresponding considerations lead to the other two formulas HaHb= c * cos (gamma) and HbHc= a * cos (alpha).

Fagnano's problem
......Of all triangles that can be placed in any triangle, the height base triangle has the smallest circumference.

More can be found at Eric W. Weisstein (MathWorld) under the keyword Fagnano's problem (Url below)

Double height base triangle Top

......If the heights are lengthened so that they encompass the circumference of the starting triangle at points Ha', Hb'and Hc'intersect and connect the intersection points, this creates the "double height base triangle".

It is called so because its sides are twice as long as those of the height base triangle. In addition, respective sides are parallel.

The points AHc'BC form a quadrilateral tendon, for which the String set applies:
AHc*HcB = CHc*HcHc'or with the designations from above pq = rHcHc'. 
From this it follows that HcHc'= (pq) / r = h. So HH appliesc= HcHc'.
HH applies accordinglyb= HbHb'and HHa= HaHa'.

After reversing the first theorem of rays, the sides are parallel.
According to the second theorem of rays, the sides of the double height base triangle twice as long as the sides of the Height base triangle.

A centric stretch Top

......A deeper insight into the interrelationships can be obtained if one looks at the centric extension of the central triangle on the given triangle.
The center of the bisector or median line is the center.
The stretch factor is k = -2.

......In this figure, the triangle SM also goescM into the triangle CHS over.
From this follows the equation HS = 2 * SM, which has already been proven above.
The following statement also applies: In a triangle, the height intersection point has twice the distance from a corner point as the center point from the side center point opposite the corner point.

The use of centric stretching is the usual and also the elegant method of examining the Euler straight line, the Feuerbach circle and the height base triangle. I had the ambition to derive the somewhat trickier sentences as far as possible using the simple methods of coordinate geometry, to tread the arduous "donkey's path" so to speak.
Quote: "The proof can be made elegantly in different ways, if necessary also by calculating with coordinates." (http://eddy.uni-duisburg.de/treitz/denkmnu/routh/cev71.htm)

Heights in the triangle on the Internet Top


Arndt Brünner
Heron's formula

Darij Grinberg
About some sentences and exercises from triangle geometry (.pdf file)

Eckart Specht

Cyril Hertz
Proof of Euler's line

Hans-Gert Gränke
Euler's straight line and Feuerbach's circle (.pdf file)

Height (geometry), height intersection, height base triangle, Taylor circle, Feuerbach circle, Euler line, Euler's theorem (geometry)


Aarnout Brombacher
An exploration of triangles and points of concurrency

Antonio Gutierrez (Go Geometry)
Example: Problem 136. Orthic Triangle, Altitudes, Orthocenter, Incenter, Perpendicular, Concyclic Points ...

Eric W. Weisstein (MathWorld)
Altitude, Orthocenter, Orthic Triangle, Euler Line, Nine-Point Circle, Taylor Circle, Fagnanos Problem,
Pedal Triangle, Euler Triangle, Euler Points, Feuerbach's Theorem, Circum-Orthic Triangle, Orthocentric System

John Page
Orthocenter of a Triangle, Geometry Construction

Pat Ballew
Orthocenter of a Triangle

The Wolfram Demonstrations Project
Example: Bisectors of the Angles of the Orthic Triangle ...

Tom Davis
Four Points on a Circle (.pdf file)

Orthocenter, Altitude (triangle), Nine-point circle, Euler line

credentials Top
Jan Gullberg: Mathematics- From the Birth of Numbers, New York, London 1997 (ISBN0-393-04002-X)

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© 2010 Jürgen Köller