Sides of orthic triangle are parallel to tangent through opposite vertex
$begingroup$
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.)
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
asked Nov 26 '18 at 3:20
SaulpilaSaulpila
1387
$endgroup$
add a comment |
$begingroup$
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.)
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
asked Nov 26 '18 at 3:20
SaulpilaSaulpila
1387
$endgroup$
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.)
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– 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
add a comment |
$begingroup$
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.)
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
asked Nov 26 '18 at 3:20
SaulpilaSaulpila
1387
$endgroup$
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.)
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
asked Nov 26 '18 at 3:20
SaulpilaSaulpila
1387
asked Nov 26 '18 at 3:20
SaulpilaSaulpila
1387
asked Nov 26 '18 at 3:20
SaulpilaSaulpila
1387
asked Nov 26 '18 at 3:20
SaulpilaSaulpila
1387
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
add a comment |
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
add a comment |
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 $triangle ABC$.
answered Nov 26 '18 at 5:25
darij grinbergdarij grinberg
10.5k33062
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add a comment |
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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 $triangle ABC$.
answered Nov 26 '18 at 5:25
darij grinbergdarij grinberg
10.5k33062
$endgroup$
add a comment |
$begingroup$
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 $triangle ABC$.
answered Nov 26 '18 at 5:25
darij grinbergdarij grinberg
10.5k33062
$endgroup$
add a comment |
$begingroup$
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 $triangle ABC$.
answered Nov 26 '18 at 5:25
darij grinbergdarij grinberg
10.5k33062
$endgroup$
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 $triangle ABC$.
answered Nov 26 '18 at 5:25
darij grinbergdarij grinberg
10.5k33062
answered Nov 26 '18 at 5:25
darij grinbergdarij grinberg
10.5k33062
answered Nov 26 '18 at 5:25
darij grinbergdarij grinberg
10.5k33062
answered Nov 26 '18 at 5:25
darij grinbergdarij grinberg
10.5k33062
10.5k33062
add a comment |
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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