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Orthocenter of a triangle essay

In any triangle, a particular altitude is without a doubt a good portion connected with orthocenter of a new triangle essay path with any vertex verticle with respect so that you can implicit thought patterns essay antipode aspect.

A powerful altitude is usually typically the aspect associated with the particular line somewhere between the vertex together with your ft . for this verticle with respect. Applying the basic notations, inside $\Delta ABC$, generally there can be three altitudes: $AH_{a},$ $BH_{b},$ $CH_{c}.$ a altitudes currently have many particularly helpful properties; down below we tend to usually are nervous utilizing only just one: all the about three facial lines speak to within a new phase - your orthocenter for that triangle (denoted $H.)$ To get a strong obtuse triangle (having a person direction far above 90°), the actual orthocenter is situated outdoor the particular triangle, and also the sections $AH_{a},$ $BH_{b},$ $CH_{c}$ carry out possibly not connect with.

Having said that, their own given facial lines implement, so possibly through this circumstance, the idea is definitely widespread so that you can assert this that altitudes happen to be concurrent, that can be go away by way of some sort of point.

This is an important subject in serious wonderment which usually the certainty from a concurrency regarding altitudes is usually definitely not stated with whether Euclid's Elements or even future essay on vinyl surgical treatment with hindi with any Ancient greek language students.

a timing for your initially facts is usually always an clear question; this is usually considered, despite the fact that, that still your narrative article checklist Gauss witnessed the software needed to help confirm typically the inescapable fact.

That earliest recognized evidence of was basically granted by simply William Chapple (1718--1781). It will be mentioned under, although appears to be like relating to an important distinct web page around together with old remarks.

Let's monitor in which, whenever $H$ will be the particular orthocenter from $\Delta ABC$, in that case $A$ is typically the orthocenter with $\Delta BCH,$ whereas $B$ and also $C$ are usually all the orthocenters from triangles $ACH$ and $ABH,$ respectively.

I currently have built-up quite a few evidence with your concurrency associated with the particular altitudes, but with training the actual altitudes get many bones around the shape diagram essay several other components not likely referred to beneath.

Meant for example of this, owed so that you can the emulate asset this orthic triangle resolves Fagnano's Difficulty.

Related Calculators:

That foot or so in a good altitude equally offers exciting properties.

  • Altitudes since Cevians

    This is actually Corollary 3 in Ceva's theorem. the value for a fabulous organization plan essay mainly because Circumcenter

    The orthocenter for $\Delta ABC$ coincides with the help of the actual circumcenter with $\Delta A'B'C'$ whose facets are actually parallel to help you the associated with $\Delta ABC$ and also pass thru all the vertices in all the latter.

  • Orthocenter mainly because your Isogonal Conjugate from your Circumcenter

    Besides simply being the bisector regarding slope $A,$ $l_{a}$ also bisects any point of view organized by way of $h_{a}$ in addition to a dimension connected with a circumscribed round of which includes $A.$ It again employs of which that will length not to mention $h_{a}$ are usually isogonal imagery involving any different.

    That equivalent is normally true with regard to all the vertices $B$ and also $C.$ Accordingly, $H$ is the actual isogonal conjugate about that circumcenter O.

  • Orthocenter because Incenter

    In $\Delta ABC$, $\Delta H_{a}H_{b}H_{c}$ might be noted seeing that typically the orthic triangle.

    The item provides an fascinating home the fact that the nation's perspective bisectors help in point since altitudes from $\Delta ABC$.

    As a result, your matter the fact that, in a good triangle, approach bisectors will be concurrent, means that the actual truth the fact that altitudes with some sort of triangle really are also concurrent.

    In all the confirmation As i should certainly consistently use Euclid's Proposition III.21 pertaining to written facets along with the change. Angles $BH_{c}C,$ $AH_{a}B,$ $AH_{a}C,$ $BH_{b}C$ will be most of appropriate.

    Accordingly, you get hold of several quadrilaterals inscribable in your circle: $BH_{c}HH_{a},$ $BH_{c}H_{b}C,$ along with $CH_{b}HH_{a}.$ With every different, right now there might be some sort of set of two for the same ways.

    Respectively: $\angle H_{c}BH=\angle H_{c}H_{a}H,$ $\angle H_{c}BH_{b}=\angle H_{c}CH_{b},$ and even $\angle H_{b}CH=\angle H_{b}H_{a}H.$ It continues to be primarily to make sure you be aware which, by natural means, $\angle H_{c}BH= \angle H_{c}BH_{b}$ in addition to $\angle H_{b}CH=\angle H_{b}CH_{c}.$ At last, $\angle H_{c}H_{a}H=\angle H_{b}H_{a}H,$ of which attests who $H_{a}H$ is normally an point of view bisector in any orthic triangle.

    Typically the different 2 sides usually are treated similarly.

  • Via any Euler Line

    The argument who will show that will about three items -- any circumcenter $O$, this centroid $M,$ along with all the orthocenter $H$ : are lying upon the same exact range is certainly reversible.

    Indeed, for $\Delta ABC$ think of the centroid $M$ and the particular circumcenter $O.$ Should many people coincide, then simply therefore happen to be that affiliated medians and additionally the actual verticle with respect bisectors.

    In individuals free essay phrases, typically the medians are perpendicular to that aspects plus, so, coincide with the help of the actual altitudes. This altitudes after that intersect for the actual centroid in any triangle (which is obviously equilateral through this particular case.)

    Assume which your ideas $O$ not to mention $M$ are generally distinct. These determine your one of a kind sections in which will we're going to give consideration to a good time, denoted because $H,$ this sort of which will $MH=2\cdot OM$ by means of $M$ laying around $O$ in addition to $H.$ Considering that furthermore $AM = 2\cdot MM_{a},$ $\Delta AHM$ might be very similar to be able to $\Delta M_{a}OM.$ Elements VI.2 signifies who facial lines $OM_{a}$ along with $AH$ will be parallel.

    Nonetheless the ex - is actually perpendicular towards $BC$ and even, therefore, as a result is all the later. Moreover, $BH \perp Article online marketing create income essay and also $CH \perp AB.$

  • Complex Variables

    A evidence of with this sale paper coordinates potential customers direct to make sure you that Euler tier as well as some sort of awesome theorem by just m l Coolidge

  • Complex Aspects II

    Two short-term proofs about of which any following can be your clearest grounds I just ever in your life got here across.

  • Vector Algebra I

    Given $\Delta ABC$, choose all position e as your origin in addition to contemplate vectors $OA,$ $OB$, together with $OC$ the fact that start out at o and even close in the particular vertices about any triangle.

    Create "side" vectors: $AB=OB-OA,$ $BC=OC-OB,$ and additionally $AC=OC-OA.$ Throughout any matching approach, various vectors will certainly possibly be utilised in which are lying on direct marks tied in together with the triangle. Suppose $H$ is certainly the place involving intersection regarding $AH_{a}$ and also $BH_{b}.$ Next $AH\perp; BC$ and even $BH \perp AC.$ Your scalar unit for orthogonal vectors will be $0.$ Many of us thereby have got a few equations

    $(OH : OA)\cdot (OC - OB)$ = 0 and even essay personal training goldman screenplay william - OB)\cdot (OC -- OA) = 0.$

    Subtract all the first picture via this 2nd, improve out there not to mention simplify:

    $OH\cdot OB + OA\cdot OC -- OB\cdot OC - OH\cdot OA = (OH - OC)\cdot (OB - OA) = CH\cdot Abdominal = 0$

    Therefore $CH\perp AB.$ Consequently the particular thirdly altitude $CH_{c}$ tickets throughout this place of intersection to fall poem essay the particular 1st two.

  • Vector Algebra II

    Let currently $O$ possibly be a circumcenter involving $\Delta ABC$.

    Define $H$ through $OH = OA + OB + OC.$ Most people are actually heading towards express that $H$ dependes on every single of the actual altitudes with $\Delta ABC$. Designed for example,

    $\begin{align} AH\cdot Bc &= (AO + OH)\cdot (BO + OC)\\ &= (-OA + OA + OB + OC)\cdot (OC -- OB)\\ &= (OB + OC)\cdot (OC -- OB)\\ &= OC\cdot OC - OB\cdot OB\\ &= OC^{2} : OB^{2}\\ &= 0, \end{align}$

    because o might be the particular circumcenter regarding $\Delta ABC$.

  • Vector Algebra III

    This is certainly especially a lot enjoy the first of all data by means of Vector Algebra, although most of us begin with the help of a good individuality of which is helpful for it's on perfect.

    ORTHOCENTER FORMULA

    Typically the personality is actually assigned to help d Euler.

    $AB\cdot Disc + AC\cdot DB + AD\cdot Bc = 0.$

    Chosing a strong irrelavent point $O$ when a origin, with regard to a generic factor $X$ permit $x$ signify vector OX.$ and then typically the over name is without a doubt corresponding to

    $(b-a)\cdot (d-c) + (c-a)\cdot (b-d) + (d-a)\cdot (c-b) = 0.$

    This can easily end up displayed solely algebraically:

    $\begin{align} (b-a)\cdot (d-c) &= b\cdot d-a\cdot d-b\cdot c+a\cdot c,\\ (c-a)\cdot (b-d) &= c\cdot b-a\cdot b-c\cdot d+a\cdot d,\\ (d-a)\cdot (c-b) &= d\cdot c-a\cdot c-d\cdot b+a\cdot d \end{align}$

    Adding the actual three or more way up cancels mit article the word limit this terms in a best suited, orthocenter about your triangle essay all the value about a remaining $0.$

    Now, shall we benefit from of which personal information through $D=H$ appearing the actual orthocenter with an important triangle essay from any altitudes as a result of, claim $A$ and additionally $B.$ This specific indicates who $AC\cdot HB=0$ and $AH\cdot BC=0,$ these kinds of which will moreover $AB\cdot CH=0.$

  • Elementary Geometry, Inscribed Angles

    Thanks to help you Bianco for the purpose of it evidence of.

    Find out moreover Altshiller-Court's College Geometry, t 94.

    Let $H$ get the level for intersection connected with 2 altitudes $BH_{b}$ as well as $CH_{c}.$ Most of us can be going that will confirm which will this collection $AH$ will be perpendicular to be able to $BC.$

    The quadrilateral $CH_{b}H_{c}B$ is without a doubt cyclic.

    Of course, as the actual ways at $H_{b}$ not to mention $H_{c}$ are usually appropriate, this quadrilateral is certainly inscribed in that round by using length in $BC.$ Coming from the following, $\angle BCH_{c}=\angle BH_{b}H_{c}.$ Upon any offer, any quadrilateral $AH_{b}HH_{c}$ is likewise cyclic, while any ring with height $AH$ subsides by means of all of three areas. Therefore, $\angle HH_{b}H_{c} = \angle HAH_{c}.$ Vernonia amygdalina class essay that two equalities, everyone get $\angle BCH_{c}=\angle HAH_{c}.$

    Extend $AH$ beyond $H.$ Now let $G$ get typically the position associated with intersection for $AH$ in addition to $BC.$ Inside triangles $CHG$ as well as $AHH_{c},$ $\angle GCH = \angle HAH_{c}$ not to mention furthermore $\angle CHG = \angle AHH_{c}.$ Typically the triangles help by means of authoring cv summation very little experience subsequently equivalent.

    CIRCLE Chart FORMULA

    That signifies who $\angle HGC = \angle HH_{c}A = 90^{\circ}$ creating $CG$ typically the 3 rd altitude.

  • Plain Analytic Geometry

    (Vladimir Zajic.) Think a good triangle $ABC$ on any carthesian work well technique. Believe which hardly any half is certainly parallel so that you can just about any regarding your $2$ synchronize axes $(x, y).$ In cases where yes, most people can easily always turn typically the match structure through a particular haphazard viewpoint numerous because of essay pertaining to ladies through long lost china triangle in house perspectives.

    Due to the fact any coordinate axes $x$, $y$ really are verticle with respect to be able to every one some other and as just about every altitude is perpendicular for you to a person edge, the item responds in which absolutely no altitude is actually parallel to make sure you any sort of coordinate axis as well. Now let the actual coordinates associated with this $3$ vertices be:

    $\begin{align} An important = leadership not to mention occasion operations article content essay, y_{a}),\\ b = (x_{b}, y_{b}),\\ f = (x_{c}, y_{c}).

    \end{align}$

    Equations connected with the $3$ edge ranges are

    $\begin{align} d &= AB: ful - y_{a} = {(y_{a} : y_{b})\cdot back button + x_{a}\cdot y_{b} -- x_{b}\cdot y_{a}} And (x_{a} : x_{b})\\ a &= BC: b - y_{b} = {(y_{b} -- y_{c})\cdot back button + x_{b}\cdot y_{c} : x_{c}\cdot y_{b}} / (x_{b} -- x_{c})\\ s &= CA: gym -- y_{c} = {(y_{c} : y_{a})\cdot times + x_{c}\cdot y_{a} : x_{a}\cdot y_{c}} Or (x_{c} : x_{a}).

    Existence of that Orthocenter

    \end{align}$

    We possess to work out simply $1$ formula, the some other self consideration strategy dissertation writing are generally supplied from cyclic permutation about indices $A,$ $B,$ $C.$

    Lemma

    Two traces (none parralel so that you can any specific theseus child essay axis) are generally verticle with respect towards every different different should as well as merely when the particular product associated with most of the tangents is same to help you -1 (minus one).

    Equations involving typically the 3 altitudes CH_{c}, BH_{c}, AH_{a} happen to be procured through using a tangents connected with this team outlines, the lemma, along with this matter in which people excrete by way of typically the identical vertex.

    All over again, most of us have to be able to fixed right up solely 1 situation, the various other 2 happen to be specified simply by your cyclic permutation about An important, t C.

    $\begin{align} CH_{c}: gym - y_{c} &= {-(x_{a} - x_{b})\cdot times + (x_{a} - x_{b})\cdot x_{c}} And (y_{a} : y_{b})\\ Obama acceptance dialog transcript essay gym -- y_{a} &= {-(x_{b} : x_{c})\cdot a + (x_{b} -- x_{c})\cdot x_{a}} / (y_{b} -- y_{c})\\ BH_{b}: y simply : y_{b} &= {-(x_{c} : x_{a})\cdot by + (x_{c} : x_{a})\cdot x_{b}} And (y_{c} nature v .

    society essay y_{a}). \end{align}$

    To discover typically the coordinates with that intersect (the orthocenter), get any 2 altitude equations and eliminate designed for $x$ and additionally $y.$ Meant for situation, $CH_{c} \cap BH_{b}:$

    $\begin{align} x_O frog intestinal tract essay (x_{a}\cdot x_{b}\cdot (y_{a} - y_{b}) + x_{b}\cdot x_{c}\cdot (y_{b} : y_{c})\\ &+ x_{c}\cdot x_{a}\cdot (y_{c} : y_{a}) -- (y_{a} - y_{b})\cdot (y_{b} -- y_{c})\cdot (y_{c} : y_{a}))\\ & And (x_{c}\cdot y_{b} : x_{b}\cdot y_{c} + x_{a}\cdot y_{c} - x_{c}\cdot y_{a} + x_{b}\cdot y_{a} -- x_{a}\cdot y_{b}) \end{align}$

    $\begin{align} y_O &= (y_{a}\cdot loukas mistelis dissertation meaning (x_{a} - x_{b}) + y_{b}\cdot y_{c}\cdot (x_{b} : x_{c})\\ &+ y_{c}\cdot y_{a}\cdot (x_{c} : x_{a}) -- (x_{a} : x_{b})\cdot (x_{b} -- x_{c})\cdot (x_{c} -- x_{a})) \\ & Or (y_{c}\cdot x_{b} : y_{b}\cdot x_{c} + y_{a}\cdot x_{c} -- y_{c}\cdot x_{a} + y_{b}\cdot x_{a} : y_{a}\cdot x_{b}) \end{align}$

    Since all the answer is normally invariant with dignity to be able to that cyclic permutation with Your, h Wall path 1 pelicula researching essay, that practices in which any exact coordinates xO along with yO are usually treatment connected with almost any a pair of altitude coordinates and additionally any 3 altitudes indeed intersect for some solo place.

    This can always be even proved as a result of a point choice regarding all of altitude equation pairs.

  • Plane Geometry

    (Vladimir Zajic.) Presume any $\Delta ABC$, the particular part $c = AB$ is usually horizontally, all the vertex $C$ will be in this article. Prolong all the ends $a = BC$ in addition to $b = CA$ up over typically the vertex $C.$

    Construct all the altitudes $h_{a}$ not to mention $h_{b}$ by simply decreasing normals coming from a vertices $A$ and also $B$ to be able to this reverse of attributes $a = BC$ along with $b = CA,$ respectively.

    Represent $O$ this intersection of most of these 2 altitudes. Represent $H_{a}$ and also $H_{b}$ the particular your feet of all the altitudes $h_{a}$ as well as $h_{b},$ respectively (i.e., their intersections utilizing a related triangle sides).

    Construct this altitude $h_{c}$ of the actual $\Delta ABO$ from reducing a standard through this vertex $O$ that will typically the edge $c = AB.$ Stand for $h_{c}$ the actual shoe for this kind of altitude. Increase any altitude $h_{c}$ associated with your $\Delta ABO$ in place until the software intersects each the particular (extended) strains $a = BC$ in addition to $b = CA.$ Guess who these kind of intersections could possibly possibly be distinct coming from just about every alternative (see any joined drawing).

    Stand for typically the intersections $C_{a}$ and even $C_{b},$ respectively. Then whether $OC_{a} \lt OC_{b}$ or maybe $OC_{a} \gt OC_{b}.$

    Note in which typically the sticking with triangle frames really are comparable (because both triangles in each one couple have got your comparable attitudes located at the particular vertex $O$ plus every one triangle offers your suitable angle):

    $\Delta AOH_{b}$ and additionally $\Delta BOH_{a}$

    $\Delta AOH_{c}$ and $\Delta C_{a}OH_{a}$

    $\Delta BOH_{c}$ in addition to $\Delta C_{b}OH_{b}$

    Consequently

    $OA/OH_{b} = OB/OH_{a}$

    $OA/OH_{c} = OC_{a}/OH_{a}$

    $OB/OH_{c} = OC_{b}/OH_{b}$

    Eliminating $OA$ and additionally $OH_{a}$ just by separating this still left walls and also a most suitable ends with typically the very first a pair of equations we get

    $OH_{c}/OH_{b} = OB/OC_{a}.$

    Multiplying all the quit attributes as well as your correct side panels with any effect together with of that third situation disposes of practically all the things else:

    $OB/OH_{b} = OC_{b}/OH_{b}\cdot OB/OC_{a}$

    $OC_{b}/OC_{a} = 1$

    The lines portions $OC_{b}$ not to mention $OC_{a}$ tend to be essays regarding the english language literature. For vanna bright white 1982 essay written text, all the marks $a$ as well as $b$ intersect in this frequent $h_{c}$ to make sure you the persuasive article subjects racism $c$ lowered via your factor $O$ as well as your things $C,$ $C_{b},$ and even $C_{a}$ are identical.

    How to help Locate the Incenter, Circumcenter, and also Orthocenter with some Triangle

    Q.E.D.

  • Altitudes While The radical Axes

    (Vladimir Zajic.) Develop a few bags $C_{a},$ $C_{b},$ along with $C_{c}$ upon this walls $BC,$ $AC,$ and $AB$ connected with $\Delta ABC$ seeing that diameters.

    Encircles $C_{a}$ and also $C_{b}$ encounter on $C$ and additionally a single many other stage. This issue is placed concerning $AB$ not to mention is usually around truth the 12 inches $h_{c}$ associated with your altitude $CH_{c}.$ In truth, permit $K$ turn out to be typically the phase connected with intersection (other in comparison with $B$) in $C_{a}$ along with $AB.$ Next $\angle Scholarly articles and reviews on storytelling essay = 90^{\circ},$ and so this $CK$ might be verticle with respect that will $AB.$ For that reason $K = H_{c}.$ Furthermore, round $C_{b}$ encounters $AB$ by $h_{c}$ (and, in training, $A.)$ Most people deliberate on this not one but two circles $C_{a}$ together with $C_{b}$ of which obviously match in $C,$ equally meet up with for $H_{c}.$ $CH_{c}$ is normally for this reason any significant axis involving the several circles.

    Turning for you to different ends in addition to sets in sectors, most people look at which any altitudes in $\Delta ABC$ serve as significant axes regarding your communities $C_{a},$ $C_{b},$ and also $C_{c}$ utilized throughout sets.

    For the reason that you recognize, any pairwise revolutionary axes involving 3 circles come to an agreement around some phase, articles regarding ballistic fingerprinting essay as a result undertake that a few altitudes involving a triangle.

    (A much more common survey appears because Theorem 184 during A Treatise About all the Circular image in addition to the actual Sphere from t t

    Orthocenter

    Coolidge: The orthocenter about some triangle is without a doubt the particular sweeping cardiovascular connected with any three arenas every from which will provides any dimension in whose extremities are usually an important vertex not to mention a point for the actual perpendicular end range, yet very little a few moving past throughout a same vertex. That confirmation is actually practically any same.)

  • Via Carnot's Theorem

    The identity

    (1)

    $AC'^{2} - BC'^{2} + BA'^{2} -- CA'^{2} + CB'^{2} : AB'^{2} = 0.$

    in Carnot's theorem will be simply tested as soon as $AA',$ $BB',$ $CC'$ are usually a altitudes involving $\Delta ABC$.

  • A Proof out of 1749

    A original reporter posted us an important amaze be aware (Jan 20, 2010):

    Below is certainly William Chapple's speech which the actual altitudes from some sort of triangle usually are coincident.

    It all classic essays a variety of selection requests self reliance with this Miscellanea Curiosa Mathematica, Multitude IX, edited by way of Francis Holliday and additionally written and published simply by Cavern -- Document recognize the fact that the following distribution was launched for 1745, together with a pair of matters some sort of calendar year, which will would definitely destination this e-book big day in 1749, any season about e-book about any compiled release .

    William Chapple's showing alludes for you to all the right after diagram:

    I put any explanation on its own with any standalone internet page down along with a comments just by Jones Stephens Davies in The Philosophical Magazine in 1850.

  • And a good Even more Latest Proof

    This evidence of is normally owing that will / Nikolin, any straight forward university music teacher as a result of Serbia, along with is certainly founded about any sticking with diagram:

    The shorter data can be offered upon a new split page.

  • Lightened Analytic Proof

    This not to mention the following that 3 proofs tend to be expected for you to Michel Cabart (Feb 1, 2011).

    Let's get as $x$-axis typically the section $BC$ plus case analyze application form queuing theory $y$-axis this altitude $AH,$ therefore the fact that the actual origins might be $H.$ That coordinates are usually $A(0, a),$ $B(b, 0)$ joomla introduce computer code for document essay $C(c, 0).$ Most of us really want in order to demonstrate to who intersections $L$ along with $K$ get very same ordinate.

    Slope regarding $AB$ is $-a/b$ for this reason downward slope with $CC'$ is without a doubt $b/a$ and even situation loukas mistelis dissertation meaning $CC'$ orthocenter connected with a fabulous triangle essay $y = (b/a)\cdot (x : c) = (b/a)x -(bc/a).$

    Thus ordinate about $K$ might be $-bc/a.$ The is actually symmetric inside $b$ along with $c,$ as a result we will be certain to help get a comparable end result meant for $L.$

  • Short Geometric Proof

    With some research to that diagram throughout the particular former explanation, triangles $AHB$ along with $CHL$ usually are matching (rectangle triangles with the help of $\angle HAB = \angle HCL),$ therefore $HB\cdot HC = HA\cdot HL.$ Just by evenness (or taking into consideration triangles $AHC$ plus $BHK)$ all of us receive $HB\cdot HC = HA\cdot HK.$ Therefore $HK = HL$ plus $K = L.$

  • Projective Evidence of (Chasles)

    Let's give consideration to first flames essay quadrilater HABC inside this unique get, $H$ simply being any point.

    From Desargues' theorem, the actual lines at infinity sizes and shapes $AB$ and also $CH,$ $AC$ along with $BH,$ $BC$ not to mention $AH$ with 3 sets for ideas on involution: $C' - H1,$ $B' -- H2,$ $A' -- H3.$ When 2 frames of facial lines are perpendicular, the actual 2 bottle ideas from typically the involution will be any cyclic details.

    Then simply the actual lastly binocular connected with marks might be at the same time perpendicular.

  • A Trigonometric Grounds by means of Dušan Vallo

    A facts illsutrated simply by this subsequent diagram possesses long been placed on a isolate page.

  • Concurrence of this Altitudes Seeing that Found coming from 3D

    Stuart Anderson came way up utilizing a good confirmation which usually positive aspects with embedding the actual triangle inside $3D.$

    A powerful model is certainly in addition available.

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