Sides of orthic triangle are parallel to tangent through opposite vertex
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 at 3:20
Saulpila
1387
add a comment |
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 at 3:20
Saulpila
1387
1
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.)
– darij grinberg
Nov 26 at 3:39
Of course, thanks! You should post it as an answer so I can accept it :).
– Saulpila
Nov 26 at 5:02
add a comment |
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 at 3:20
Saulpila
1387
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 at 3:20
Saulpila
1387
asked Nov 26 at 3:20
Saulpila
1387
asked Nov 26 at 3:20
Saulpila
1387
asked Nov 26 at 3:20
Saulpila
1387
asked Nov 26 at 3:20
Saulpila
1387
1387
1
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.)
– darij grinberg
Nov 26 at 3:39
Of course, thanks! You should post it as an answer so I can accept it :).
– Saulpila
Nov 26 at 5:02
add a comment |
1
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.)
– darij grinberg
Nov 26 at 3:39
Of course, thanks! You should post it as an answer so I can accept it :).
– Saulpila
Nov 26 at 5:02
1
1
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.)
– darij grinberg
Nov 26 at 3:39
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.)
– darij grinberg
Nov 26 at 3:39
Of course, thanks! You should post it as an answer so I can accept it :).
– Saulpila
Nov 26 at 5:02
Of course, thanks! You should post it as an answer so I can accept it :).
– Saulpila
Nov 26 at 5:02
add a comment |
1 Answer
1
active
oldest
votes
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 at 5:25
darij grinberg
10.2k33061
add a comment |
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
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
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
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 at 5:25
darij grinberg
10.2k33061
add a comment |
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 at 5:25
darij grinberg
10.2k33061
add a comment |
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 at 5:25
darij grinberg
10.2k33061
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 at 5:25
darij grinberg
10.2k33061
answered Nov 26 at 5:25
darij grinberg
10.2k33061
answered Nov 26 at 5:25
darij grinberg
10.2k33061
answered Nov 26 at 5:25
darij grinberg
10.2k33061
10.2k33061
add a comment |
add a comment |
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
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
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
1
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.)
– darij grinberg
Nov 26 at 3:39
Of course, thanks! You should post it as an answer so I can accept it :).
– Saulpila
Nov 26 at 5:02