Sides of orthic triangle are parallel to tangent through opposite vertex












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$begingroup$


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







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    $begingroup$
    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






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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
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$begingroup$

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 $triangle ABC$.






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    $begingroup$

    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

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