# 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 h_{c}, divides a triangle into two right triangles.In the right triangle, h applies _{c}= a * sin (beta), in the left h_{c}= b * sin (alpha).So there are two ways to change the height h _{c} to calculate.The following two formulas are derived accordingly, H _{a}= b * sin (gamma) = c * sin (beta) and h_{b}= c * sin (alpha) = a * sin (gamma) |

Sine law

From h

_{c}= a * sin (beta) and h

_{c}= 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 H _{a}= (2 / a) sqrt [s (s-a) (s-b) (s-c)],H _{b}= (2 / b) sqrt [s (s-a) (s-b) (s-c)],H _{c}= (2 / c) sqrt [s (s-a) (s-b) (s-c)].Where s = (1/2) (a + b + c). |

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) ch

_{c}and solves the Heron formula according to h

_{c}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. |

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.

**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) ch |

A new formula results from the area formula.

...... | The relationship is ah_{a} = bra_{b }= ch_{c}. 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) ah _{a}, A = (1/2) bh_{b} and A = (1/2) ch_{c} is 1 / r = 1 / h_{a} + 1 / h_{b}+ 1 / h_{c}.That's another formula. |

**Elevation sections **Top

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

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 AHH_{c} and CHH_{a}. The proportion CH applies to the marked sides |

_{b}with one, it is CH * HH

_{c}= AH * HH

_{a}= BH * HH

_{b}.

The product of the two height sections of a height is constant.

**Euler straight line **Top

sentence

...... | 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 h_{c}= r and that by the height h_{c} 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 M_{a}(p / 2 | r / 2) and M_{c}[(p-q) / 2 | 0].The x coordinate of M _{c} results from (p + q) / 2-q = (p-q) / 2.The straight line MM _{a} stands vertically on side a and therefore has the slope p / r. She goes through point M_{c} and therefore has with the approach (y-y_{1}) = m (x-x_{1}) the equation (y-r / 2) = (p / r) (x-p / 2) or y = (p / r) x-p² / 2r + r / 2. |

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. |

Collinearity

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.

_{1}) / (x-x

_{1}) = (y

_{2}-y

_{1}) / (x

_{2}-x

_{1})

<=> (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

This means that S is also on HM.

What remains is HS = 2 * SM.

The lengths of the lines HM and SH are calculated using the formula s = sqrt [(x

_{2}-x

_{1}) ² + (y

_{2}-y

_{1}) 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}, H_{b}, H_{c}, the midpoints of the sides M_{a}, M_{b}, M_{c} and the midpoints of the height sections H emanating from the corner points_{1}, H_{2} and H_{3}. 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 M |

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 = x

_{1}+ (1/2) (x

_{2}-x

_{1}) and y = y

_{1}+ (1/2) (y

_{2}-y

_{1}), which for the midpoint of a segment P

_{1}P.

_{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)]

radius

The radius of the Feuerbach circle is H.

_{C.}F and is based on the formula P

_{1}P.

_{2}== sqrt [(x

_{2}-x

_{1}) ² + (y

_{2}-y

_{1}) ²] calculated.

With H

_{c}(0 | 0) is H.

_{c}F² = [(1/4) (p-q)] ² + [(1/4) (h + r)] ².

Circular equation

The circle equation is calculated using the formula (x-x

_{m}) 2 + (y-y

_{m}) ² = r², where the center point M (x

_{m}| y

_{m}) 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 M_{c} , H_{c}, and H_{3} lie on the circle.

H

_{c}(0|0)

If one sets x = y = 0, then the same terms result on the left and right. The circle equation is fulfilled.

M.

_{c}

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.

H

_{3}

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 H

_{c}F² = [(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 [(x

_{2}-x

_{1}) ² + (y

_{2}-y

_{1}) 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

definition

...... | 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 AH_{c}HH_{b} is a chordal quadrilateral because the interior angles at the opposite points H_{b} and H_{c} 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 HH_{b} are. |

_{a}equal.

...... | Consider the right triangles AHH_{b} and HBH_{a}. Because of the equal vertex angles, they are similar. So the blue and red angles are the same size. So the angle becomes H _{b}H_{c}H_{a} by HH_{c} halved.In the same way one shows that the other two interior angles at H |

Interior angle

The interior angles of the height base triangle are 180 ° -2 * alpha, 180 ° -2 * beta, and 180 ° -2 * gamma.

proof

...... | The straight line AH_{a} is the height of the triangle and at the same time the bisector of the angle H_{b}H_{a}H_{c}. 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'. |

...... | 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. |

pages

The formulas H apply to the sides

_{a}H

_{c}= b * cos (beta), H

_{a}H

_{b}= c * cos (gamma) and H

_{b}H

_{c}= a * cos (alpha).

proof

It should be the side x = H

_{c}H

_{a}according to the formula s = sqrt [(x

_{2}-x

_{1}) ² + (y

_{2}-y

_{1}) ²].

...... | For this you need the coordinates of H_{c} and H_{a}. It is H_{c}(0|0).To calculate the coordinates of H _{a} bring the straight line AH_{a} and BC to the cut.AH _{a}: y = (p / r) x + h, where h = (pq) / rThe following applies: y = - (r / p) x + r = - (r / p) (pr²-p²q) / (p² + r²) + r = (pqr + p²r) / (p² + r²). |

With a² = p² + r² and b²q² + r² and p = a * cos (beta), s² = (p² / a

^{4}) (a²b²) = (p² / a²) b² or s = (p / a) b.

With p = a * cos (beta), s = b * cos (beta) = H

_{a}H

_{c}.

Corresponding considerations lead to the other two formulas H

_{a}H

_{b}= c * cos (gamma) and H

_{b}H

_{c}= 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 H_{a}', H_{b}'and H_{c}'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 AH

_{c}'BC form a quadrilateral tendon, for which the

__String set__applies:

AH

_{c}*H

_{c}B = CH

_{c}*H

_{c}H

_{c}'or with the designations from above pq = rH

_{c}H

_{c}'.

From this it follows that H

_{c}H

_{c}'= (pq) / r = h. So HH applies

_{c}= H

_{c}H

_{c}'.

HH applies accordingly

_{b}= H

_{b}H

_{b}'and HH

_{a}= H

_{a}H

_{a}'.

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 goes_{c}M 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. |

comment

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

German

Arndt Brünner

Heron's formula

Darij Grinberg

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

Eckart Specht

Math4you

Cyril Hertz

Proof of Euler's line

Hans-Gert Gränke

Euler's straight line and Feuerbach's circle (.pdf file)

Wikipedia

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

English

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)

Wikipedia

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)

**Feedback:**Email address on my main page

URL of my homepage:

http://www.mathematische-basteleien.de/

© 2010 Jürgen Köller

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