Sides of orthic triangle are parallel to tangent through opposite vertex












1












$begingroup$


Let $ABC$ be a triangle with altitudes $AD$, $BE$, $CF$. Prove that $EF$ is parallel to the tangent to the circumcircle of $ABC$ at $A$. (And similarly $DF$ and $DE$ are parallel to the tangents through $B$ and $C$, respectively.)Diagram



I'm trying to find a simple direct proof for this, but I can't seem to come up with one.






geometry triangle tangent-line







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Just check that the tangent to the circumcircle of $triangle ABC$ at $A$ intersects $AB$ at the same angle as the line $EF$. (The former angle is obtained from the chordal-angle theorem; the latter from the inscribed quadrilateral $AEHF$, where $H$ is the orthocenter. Google "antiparallels" for more about this.)
    $endgroup$
    – darij grinberg
    Nov 26 '18 at 3:39










  • $begingroup$
    Of course, thanks! You should post it as an answer so I can accept it :).
    $endgroup$
    – Saulpila
    Nov 26 '18 at 5:02
















1












$begingroup$


Let $ABC$ be a triangle with altitudes $AD$, $BE$, $CF$. Prove that $EF$ is parallel to the tangent to the circumcircle of $ABC$ at $A$. (And similarly $DF$ and $DE$ are parallel to the tangents through $B$ and $C$, respectively.)Diagram



I'm trying to find a simple direct proof for this, but I can't seem to come up with one.






geometry triangle tangent-line







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Just check that the tangent to the circumcircle of $triangle ABC$ at $A$ intersects $AB$ at the same angle as the line $EF$. (The former angle is obtained from the chordal-angle theorem; the latter from the inscribed quadrilateral $AEHF$, where $H$ is the orthocenter. Google "antiparallels" for more about this.)
    $endgroup$
    – darij grinberg
    Nov 26 '18 at 3:39










  • $begingroup$
    Of course, thanks! You should post it as an answer so I can accept it :).
    $endgroup$
    – Saulpila
    Nov 26 '18 at 5:02














1












1








1





$begingroup$


Let $ABC$ be a triangle with altitudes $AD$, $BE$, $CF$. Prove that $EF$ is parallel to the tangent to the circumcircle of $ABC$ at $A$. (And similarly $DF$ and $DE$ are parallel to the tangents through $B$ and $C$, respectively.)Diagram



I'm trying to find a simple direct proof for this, but I can't seem to come up with one.






geometry triangle tangent-line







share|cite|improve this question









$endgroup$




Let $ABC$ be a triangle with altitudes $AD$, $BE$, $CF$. Prove that $EF$ is parallel to the tangent to the circumcircle of $ABC$ at $A$. (And similarly $DF$ and $DE$ are parallel to the tangents through $B$ and $C$, respectively.)Diagram



I'm trying to find a simple direct proof for this, but I can't seem to come up with one.






geometry triangle tangent-line




geometry triangle tangent-line






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 26 '18 at 3:20









SaulpilaSaulpila

1387




1387








  • 1




    $begingroup$
    Just check that the tangent to the circumcircle of $triangle ABC$ at $A$ intersects $AB$ at the same angle as the line $EF$. (The former angle is obtained from the chordal-angle theorem; the latter from the inscribed quadrilateral $AEHF$, where $H$ is the orthocenter. Google "antiparallels" for more about this.)
    $endgroup$
    – darij grinberg
    Nov 26 '18 at 3:39










  • $begingroup$
    Of course, thanks! You should post it as an answer so I can accept it :).
    $endgroup$
    – Saulpila
    Nov 26 '18 at 5:02














  • 1




    $begingroup$
    Just check that the tangent to the circumcircle of $triangle ABC$ at $A$ intersects $AB$ at the same angle as the line $EF$. (The former angle is obtained from the chordal-angle theorem; the latter from the inscribed quadrilateral $AEHF$, where $H$ is the orthocenter. Google "antiparallels" for more about this.)
    $endgroup$
    – darij grinberg
    Nov 26 '18 at 3:39










  • $begingroup$
    Of course, thanks! You should post it as an answer so I can accept it :).
    $endgroup$
    – Saulpila
    Nov 26 '18 at 5:02








1




1




$begingroup$
Just check that the tangent to the circumcircle of $triangle ABC$ at $A$ intersects $AB$ at the same angle as the line $EF$. (The former angle is obtained from the chordal-angle theorem; the latter from the inscribed quadrilateral $AEHF$, where $H$ is the orthocenter. Google "antiparallels" for more about this.)
$endgroup$
– darij grinberg
Nov 26 '18 at 3:39




$begingroup$
Just check that the tangent to the circumcircle of $triangle ABC$ at $A$ intersects $AB$ at the same angle as the line $EF$. (The former angle is obtained from the chordal-angle theorem; the latter from the inscribed quadrilateral $AEHF$, where $H$ is the orthocenter. Google "antiparallels" for more about this.)
$endgroup$
– darij grinberg
Nov 26 '18 at 3:39












$begingroup$
Of course, thanks! You should post it as an answer so I can accept it :).
$endgroup$
– Saulpila
Nov 26 '18 at 5:02




$begingroup$
Of course, thanks! You should post it as an answer so I can accept it :).
$endgroup$
– Saulpila
Nov 26 '18 at 5:02










1 Answer
1






active

oldest

votes


















1












$begingroup$

We shall use directed angles modulo $180^{circ}$.



Let $x$ be the tangent to the circumcircle of $triangle ABC$ at $A$. Thus, $measuredangle left(AC, xright) = measuredangle CBA$ (by the well-known theorem stating that a tangent-chord angle equals the corresponding chordal angle).



Let $H$ be the orthocenter of $triangle ABC$. The points $E$ and $F$ lie on the circle with diameter $AH$ (since both angles $measuredangle AEH$ and $measuredangle AFH$ are $90^{circ}$). Thus, the points $A$, $E$, $F$ and $H$ lie on one circle. Hence, the chordal angle theorem yields
begin{align}
measuredangle AEF
& = measuredangle AHF = measuredangle left(AH, HCright)
= underbrace{measuredangle left(AH, BCright)}_{=90^circ text{ (since $AH perp BC$)}}
+ underbrace{measuredangle left(BC, ABright)}_{=measuredangle CBA}
+ underbrace{measuredangle left(AB, HCright)}_{=90^circ text{ (since $HC perp AB$)}} \
&= 90^circ + measuredangle CBA + 90^circ = underbrace{180^circ}_{=0^circ} + measuredangle CBA = measuredangle CBA .
end{align}
Hence, $measuredangle left(AC, EFright) = measuredangle AEF = measuredangle CBA = measuredangle left(AC, xright)$, so that $EF parallel x$, qed. $blacksquare$



In the language of triangle geometry, lines parallel to $x$ and $EF$ are said to be antiparallel to $BC$ with respect to $triangle ABC$.






share|cite|improve this answer









$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3013776%2fsides-of-orthic-triangle-are-parallel-to-tangent-through-opposite-vertex%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    We shall use directed angles modulo $180^{circ}$.



    Let $x$ be the tangent to the circumcircle of $triangle ABC$ at $A$. Thus, $measuredangle left(AC, xright) = measuredangle CBA$ (by the well-known theorem stating that a tangent-chord angle equals the corresponding chordal angle).



    Let $H$ be the orthocenter of $triangle ABC$. The points $E$ and $F$ lie on the circle with diameter $AH$ (since both angles $measuredangle AEH$ and $measuredangle AFH$ are $90^{circ}$). Thus, the points $A$, $E$, $F$ and $H$ lie on one circle. Hence, the chordal angle theorem yields
    begin{align}
    measuredangle AEF
    & = measuredangle AHF = measuredangle left(AH, HCright)
    = underbrace{measuredangle left(AH, BCright)}_{=90^circ text{ (since $AH perp BC$)}}
    + underbrace{measuredangle left(BC, ABright)}_{=measuredangle CBA}
    + underbrace{measuredangle left(AB, HCright)}_{=90^circ text{ (since $HC perp AB$)}} \
    &= 90^circ + measuredangle CBA + 90^circ = underbrace{180^circ}_{=0^circ} + measuredangle CBA = measuredangle CBA .
    end{align}
    Hence, $measuredangle left(AC, EFright) = measuredangle AEF = measuredangle CBA = measuredangle left(AC, xright)$, so that $EF parallel x$, qed. $blacksquare$



    In the language of triangle geometry, lines parallel to $x$ and $EF$ are said to be antiparallel to $BC$ with respect to $triangle ABC$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      We shall use directed angles modulo $180^{circ}$.



      Let $x$ be the tangent to the circumcircle of $triangle ABC$ at $A$. Thus, $measuredangle left(AC, xright) = measuredangle CBA$ (by the well-known theorem stating that a tangent-chord angle equals the corresponding chordal angle).



      Let $H$ be the orthocenter of $triangle ABC$. The points $E$ and $F$ lie on the circle with diameter $AH$ (since both angles $measuredangle AEH$ and $measuredangle AFH$ are $90^{circ}$). Thus, the points $A$, $E$, $F$ and $H$ lie on one circle. Hence, the chordal angle theorem yields
      begin{align}
      measuredangle AEF
      & = measuredangle AHF = measuredangle left(AH, HCright)
      = underbrace{measuredangle left(AH, BCright)}_{=90^circ text{ (since $AH perp BC$)}}
      + underbrace{measuredangle left(BC, ABright)}_{=measuredangle CBA}
      + underbrace{measuredangle left(AB, HCright)}_{=90^circ text{ (since $HC perp AB$)}} \
      &= 90^circ + measuredangle CBA + 90^circ = underbrace{180^circ}_{=0^circ} + measuredangle CBA = measuredangle CBA .
      end{align}
      Hence, $measuredangle left(AC, EFright) = measuredangle AEF = measuredangle CBA = measuredangle left(AC, xright)$, so that $EF parallel x$, qed. $blacksquare$



      In the language of triangle geometry, lines parallel to $x$ and $EF$ are said to be antiparallel to $BC$ with respect to $triangle ABC$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        We shall use directed angles modulo $180^{circ}$.



        Let $x$ be the tangent to the circumcircle of $triangle ABC$ at $A$. Thus, $measuredangle left(AC, xright) = measuredangle CBA$ (by the well-known theorem stating that a tangent-chord angle equals the corresponding chordal angle).



        Let $H$ be the orthocenter of $triangle ABC$. The points $E$ and $F$ lie on the circle with diameter $AH$ (since both angles $measuredangle AEH$ and $measuredangle AFH$ are $90^{circ}$). Thus, the points $A$, $E$, $F$ and $H$ lie on one circle. Hence, the chordal angle theorem yields
        begin{align}
        measuredangle AEF
        & = measuredangle AHF = measuredangle left(AH, HCright)
        = underbrace{measuredangle left(AH, BCright)}_{=90^circ text{ (since $AH perp BC$)}}
        + underbrace{measuredangle left(BC, ABright)}_{=measuredangle CBA}
        + underbrace{measuredangle left(AB, HCright)}_{=90^circ text{ (since $HC perp AB$)}} \
        &= 90^circ + measuredangle CBA + 90^circ = underbrace{180^circ}_{=0^circ} + measuredangle CBA = measuredangle CBA .
        end{align}
        Hence, $measuredangle left(AC, EFright) = measuredangle AEF = measuredangle CBA = measuredangle left(AC, xright)$, so that $EF parallel x$, qed. $blacksquare$



        In the language of triangle geometry, lines parallel to $x$ and $EF$ are said to be antiparallel to $BC$ with respect to $triangle ABC$.






        share|cite|improve this answer









        $endgroup$



        We shall use directed angles modulo $180^{circ}$.



        Let $x$ be the tangent to the circumcircle of $triangle ABC$ at $A$. Thus, $measuredangle left(AC, xright) = measuredangle CBA$ (by the well-known theorem stating that a tangent-chord angle equals the corresponding chordal angle).



        Let $H$ be the orthocenter of $triangle ABC$. The points $E$ and $F$ lie on the circle with diameter $AH$ (since both angles $measuredangle AEH$ and $measuredangle AFH$ are $90^{circ}$). Thus, the points $A$, $E$, $F$ and $H$ lie on one circle. Hence, the chordal angle theorem yields
        begin{align}
        measuredangle AEF
        & = measuredangle AHF = measuredangle left(AH, HCright)
        = underbrace{measuredangle left(AH, BCright)}_{=90^circ text{ (since $AH perp BC$)}}
        + underbrace{measuredangle left(BC, ABright)}_{=measuredangle CBA}
        + underbrace{measuredangle left(AB, HCright)}_{=90^circ text{ (since $HC perp AB$)}} \
        &= 90^circ + measuredangle CBA + 90^circ = underbrace{180^circ}_{=0^circ} + measuredangle CBA = measuredangle CBA .
        end{align}
        Hence, $measuredangle left(AC, EFright) = measuredangle AEF = measuredangle CBA = measuredangle left(AC, xright)$, so that $EF parallel x$, qed. $blacksquare$



        In the language of triangle geometry, lines parallel to $x$ and $EF$ are said to be antiparallel to $BC$ with respect to $triangle ABC$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 26 '18 at 5:25









        darij grinbergdarij grinberg

        10.5k33062




        10.5k33062






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3013776%2fsides-of-orthic-triangle-are-parallel-to-tangent-through-opposite-vertex%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            How to change which sound is reproduced for terminal bell?

            Image
            3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question
            Read more

            Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

            Image
            up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx
            Read more

            Can I use Tabulator js library in my java Spring + Thymeleaf project?

            Image
            0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"
            Read more