Three Altitudes of a triangle are concurrent
$begingroup$
I have been told that this well known fact can be shown using only Euclid's propositions from books one to three, and cyclic quadrilaterals.
I can't figure out how to start, which quadrilateral should I have in consideration. Also the only proposition about cyclic quadrilaterals is 3.22, which I am not sure how to use.
Edit: I think I have made some progress. Let CE and AF be altitudes from C and A, the idea is to show that the line going through B and D is perpendicular to AC. Which could be done by 3.22 if either AGDE or GCFD were cyclic quadrilaterals. I don't know how to show this last part.
geometry euclidean-geometry triangles
asked Mar 16 '16 at 20:26
SmurfSmurf
593313
$endgroup$
add a comment |
$begingroup$
I have been told that this well known fact can be shown using only Euclid's propositions from books one to three, and cyclic quadrilaterals.
I can't figure out how to start, which quadrilateral should I have in consideration. Also the only proposition about cyclic quadrilaterals is 3.22, which I am not sure how to use.
Edit: I think I have made some progress. Let CE and AF be altitudes from C and A, the idea is to show that the line going through B and D is perpendicular to AC. Which could be done by 3.22 if either AGDE or GCFD were cyclic quadrilaterals. I don't know how to show this last part.
geometry euclidean-geometry triangles
asked Mar 16 '16 at 20:26
SmurfSmurf
593313
$endgroup$
$begingroup$
I think you mean AF and GCFD instead of AE and GCED. (The F looks like it could be an E in the diagram.)
$endgroup$
– Théophile
Mar 16 '16 at 22:38
$begingroup$
This doesn't help you, but you could do this just using formulas if you were allowed.
$endgroup$
– barrycarter
Mar 16 '16 at 23:22
add a comment |
$begingroup$
I have been told that this well known fact can be shown using only Euclid's propositions from books one to three, and cyclic quadrilaterals.
I can't figure out how to start, which quadrilateral should I have in consideration. Also the only proposition about cyclic quadrilaterals is 3.22, which I am not sure how to use.
Edit: I think I have made some progress. Let CE and AF be altitudes from C and A, the idea is to show that the line going through B and D is perpendicular to AC. Which could be done by 3.22 if either AGDE or GCFD were cyclic quadrilaterals. I don't know how to show this last part.
geometry euclidean-geometry triangles
asked Mar 16 '16 at 20:26
SmurfSmurf
593313
$endgroup$
I have been told that this well known fact can be shown using only Euclid's propositions from books one to three, and cyclic quadrilaterals.
I can't figure out how to start, which quadrilateral should I have in consideration. Also the only proposition about cyclic quadrilaterals is 3.22, which I am not sure how to use.
Edit: I think I have made some progress. Let CE and AF be altitudes from C and A, the idea is to show that the line going through B and D is perpendicular to AC. Which could be done by 3.22 if either AGDE or GCFD were cyclic quadrilaterals. I don't know how to show this last part.
geometry euclidean-geometry triangles
geometry euclidean-geometry triangles
asked Mar 16 '16 at 20:26
SmurfSmurf
593313
asked Mar 16 '16 at 20:26
SmurfSmurf
593313
edited Mar 16 '16 at 22:45
Smurf
asked Mar 16 '16 at 20:26
SmurfSmurf
593313
asked Mar 16 '16 at 20:26
SmurfSmurf
593313
asked Mar 16 '16 at 20:26
SmurfSmurf
593313
593313
$begingroup$
I think you mean AF and GCFD instead of AE and GCED. (The F looks like it could be an E in the diagram.)
$endgroup$
– Théophile
Mar 16 '16 at 22:38
$begingroup$
This doesn't help you, but you could do this just using formulas if you were allowed.
$endgroup$
– barrycarter
Mar 16 '16 at 23:22
add a comment |
$begingroup$
I think you mean AF and GCFD instead of AE and GCED. (The F looks like it could be an E in the diagram.)
$endgroup$
– Théophile
Mar 16 '16 at 22:38
$begingroup$
This doesn't help you, but you could do this just using formulas if you were allowed.
$endgroup$
– barrycarter
Mar 16 '16 at 23:22
$begingroup$
I think you mean AF and GCFD instead of AE and GCED. (The F looks like it could be an E in the diagram.)
$endgroup$
– Théophile
Mar 16 '16 at 22:38
$begingroup$
I think you mean AF and GCFD instead of AE and GCED. (The F looks like it could be an E in the diagram.)
$endgroup$
– Théophile
Mar 16 '16 at 22:38
$begingroup$
This doesn't help you, but you could do this just using formulas if you were allowed.
$endgroup$
– barrycarter
Mar 16 '16 at 23:22
$begingroup$
This doesn't help you, but you could do this just using formulas if you were allowed.
$endgroup$
– barrycarter
Mar 16 '16 at 23:22
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Since $AEFC$ is cyclic, $angle FEB =angle ACB$.
Since $BEDF$ is cyclic, $angle FEB=angle BDF$.
Thus $angle GDF+angle ACB=180^circ$.
answered Mar 17 '16 at 5:51
Quang HoangQuang Hoang
13.2k1233
$endgroup$
$begingroup$
The problem I find is to show that the quadrilaterals are cyclic from Euclid's books, since it does not seem to be any proposition about it.
$endgroup$
– Smurf
Mar 17 '16 at 7:53
$begingroup$
In this case, it's fairly easy: $A$, $E$, $F$, $C$ all lie on the circle with diameter $AC$, while $B$, $E$, $D$, $F$ lie on the circle with diameter $BD$. In general, you can check whether the two opposite angles add up to $180^circ$ (like in $BEDF$) or the two diagonal-side angles looking at a side are equal (like in $AEFC$.
$endgroup$
– Quang Hoang
Mar 17 '16 at 13:47
1
$begingroup$
Two things: first I think $angle BDE$ should read $angle BDF$ in line 2.
$endgroup$
– Eric Haney
Mar 20 '16 at 6:18
$begingroup$
Also, one way to justify that $AEFC$ is cyclic is to prove the converse of Prop 31 which says the angle in the semicircle is right. Show that a right angle standing on a straight line has its vertex fall on the semicircle described on that straight line as a diameter. In your diagram, we notice that $angle AEC$ and $angle AFC$ are right by construction. This would be sufficient to show that $AEFC$ is a cyclic quadrilateral.
$endgroup$
– Eric Haney
Mar 20 '16 at 6:27
add a comment |
$begingroup$
The easiest way I know of showing the altitudes of $ABC$ are concurrent is (1)Prove the right bisectors of a triangle are concurrent. (2) Draw line $l_A$ thru $A$ parallel to $BC$, line $l_B$ thru $B$ parallel to $CA,$ and $l_C$ thru $C$ parallel to $AB.$ These lines meet pair-wise at points $A',B',C'.$ The altitudes of $ABC$ are the right bisectors of $A'B'C'.$
answered Mar 16 '16 at 22:33
DanielWainfleetDanielWainfleet
35.7k31648
$endgroup$
$begingroup$
There is a lot to argument there. I am trying to do a step by step proof with explicit mention of the propositions used on each step. Also would like to follow the cyclic quadrilaterals path.
$endgroup$
– Smurf
Mar 16 '16 at 22:42
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%2f1700740%2fthree-altitudes-of-a-triangle-are-concurrent%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Since $AEFC$ is cyclic, $angle FEB =angle ACB$.
Since $BEDF$ is cyclic, $angle FEB=angle BDF$.
Thus $angle GDF+angle ACB=180^circ$.
answered Mar 17 '16 at 5:51
Quang HoangQuang Hoang
13.2k1233
$endgroup$
$begingroup$
The problem I find is to show that the quadrilaterals are cyclic from Euclid's books, since it does not seem to be any proposition about it.
$endgroup$
– Smurf
Mar 17 '16 at 7:53
$begingroup$
In this case, it's fairly easy: $A$, $E$, $F$, $C$ all lie on the circle with diameter $AC$, while $B$, $E$, $D$, $F$ lie on the circle with diameter $BD$. In general, you can check whether the two opposite angles add up to $180^circ$ (like in $BEDF$) or the two diagonal-side angles looking at a side are equal (like in $AEFC$.
$endgroup$
– Quang Hoang
Mar 17 '16 at 13:47
1
$begingroup$
Two things: first I think $angle BDE$ should read $angle BDF$ in line 2.
$endgroup$
– Eric Haney
Mar 20 '16 at 6:18
$begingroup$
Also, one way to justify that $AEFC$ is cyclic is to prove the converse of Prop 31 which says the angle in the semicircle is right. Show that a right angle standing on a straight line has its vertex fall on the semicircle described on that straight line as a diameter. In your diagram, we notice that $angle AEC$ and $angle AFC$ are right by construction. This would be sufficient to show that $AEFC$ is a cyclic quadrilateral.
$endgroup$
– Eric Haney
Mar 20 '16 at 6:27
add a comment |
$begingroup$
Since $AEFC$ is cyclic, $angle FEB =angle ACB$.
Since $BEDF$ is cyclic, $angle FEB=angle BDF$.
Thus $angle GDF+angle ACB=180^circ$.
answered Mar 17 '16 at 5:51
Quang HoangQuang Hoang
13.2k1233
$endgroup$
$begingroup$
The problem I find is to show that the quadrilaterals are cyclic from Euclid's books, since it does not seem to be any proposition about it.
$endgroup$
– Smurf
Mar 17 '16 at 7:53
$begingroup$
In this case, it's fairly easy: $A$, $E$, $F$, $C$ all lie on the circle with diameter $AC$, while $B$, $E$, $D$, $F$ lie on the circle with diameter $BD$. In general, you can check whether the two opposite angles add up to $180^circ$ (like in $BEDF$) or the two diagonal-side angles looking at a side are equal (like in $AEFC$.
$endgroup$
– Quang Hoang
Mar 17 '16 at 13:47
1
$begingroup$
Two things: first I think $angle BDE$ should read $angle BDF$ in line 2.
$endgroup$
– Eric Haney
Mar 20 '16 at 6:18
$begingroup$
Also, one way to justify that $AEFC$ is cyclic is to prove the converse of Prop 31 which says the angle in the semicircle is right. Show that a right angle standing on a straight line has its vertex fall on the semicircle described on that straight line as a diameter. In your diagram, we notice that $angle AEC$ and $angle AFC$ are right by construction. This would be sufficient to show that $AEFC$ is a cyclic quadrilateral.
$endgroup$
– Eric Haney
Mar 20 '16 at 6:27
add a comment |
$begingroup$
Since $AEFC$ is cyclic, $angle FEB =angle ACB$.
Since $BEDF$ is cyclic, $angle FEB=angle BDF$.
Thus $angle GDF+angle ACB=180^circ$.
answered Mar 17 '16 at 5:51
Quang HoangQuang Hoang
13.2k1233
$endgroup$
Since $AEFC$ is cyclic, $angle FEB =angle ACB$.
Since $BEDF$ is cyclic, $angle FEB=angle BDF$.
Thus $angle GDF+angle ACB=180^circ$.
answered Mar 17 '16 at 5:51
Quang HoangQuang Hoang
13.2k1233
edited Mar 20 '16 at 14:46
answered Mar 17 '16 at 5:51
Quang HoangQuang Hoang
13.2k1233
answered Mar 17 '16 at 5:51
Quang HoangQuang Hoang
13.2k1233
answered Mar 17 '16 at 5:51
Quang HoangQuang Hoang
13.2k1233
13.2k1233
$begingroup$
The problem I find is to show that the quadrilaterals are cyclic from Euclid's books, since it does not seem to be any proposition about it.
$endgroup$
– Smurf
Mar 17 '16 at 7:53
$begingroup$
In this case, it's fairly easy: $A$, $E$, $F$, $C$ all lie on the circle with diameter $AC$, while $B$, $E$, $D$, $F$ lie on the circle with diameter $BD$. In general, you can check whether the two opposite angles add up to $180^circ$ (like in $BEDF$) or the two diagonal-side angles looking at a side are equal (like in $AEFC$.
$endgroup$
– Quang Hoang
Mar 17 '16 at 13:47
1
$begingroup$
Two things: first I think $angle BDE$ should read $angle BDF$ in line 2.
$endgroup$
– Eric Haney
Mar 20 '16 at 6:18
$begingroup$
Also, one way to justify that $AEFC$ is cyclic is to prove the converse of Prop 31 which says the angle in the semicircle is right. Show that a right angle standing on a straight line has its vertex fall on the semicircle described on that straight line as a diameter. In your diagram, we notice that $angle AEC$ and $angle AFC$ are right by construction. This would be sufficient to show that $AEFC$ is a cyclic quadrilateral.
$endgroup$
– Eric Haney
Mar 20 '16 at 6:27
add a comment |
$begingroup$
The problem I find is to show that the quadrilaterals are cyclic from Euclid's books, since it does not seem to be any proposition about it.
$endgroup$
– Smurf
Mar 17 '16 at 7:53
$begingroup$
In this case, it's fairly easy: $A$, $E$, $F$, $C$ all lie on the circle with diameter $AC$, while $B$, $E$, $D$, $F$ lie on the circle with diameter $BD$. In general, you can check whether the two opposite angles add up to $180^circ$ (like in $BEDF$) or the two diagonal-side angles looking at a side are equal (like in $AEFC$.
$endgroup$
– Quang Hoang
Mar 17 '16 at 13:47
1
$begingroup$
Two things: first I think $angle BDE$ should read $angle BDF$ in line 2.
$endgroup$
– Eric Haney
Mar 20 '16 at 6:18
$begingroup$
Also, one way to justify that $AEFC$ is cyclic is to prove the converse of Prop 31 which says the angle in the semicircle is right. Show that a right angle standing on a straight line has its vertex fall on the semicircle described on that straight line as a diameter. In your diagram, we notice that $angle AEC$ and $angle AFC$ are right by construction. This would be sufficient to show that $AEFC$ is a cyclic quadrilateral.
$endgroup$
– Eric Haney
Mar 20 '16 at 6:27
$begingroup$
The problem I find is to show that the quadrilaterals are cyclic from Euclid's books, since it does not seem to be any proposition about it.
$endgroup$
– Smurf
Mar 17 '16 at 7:53
$begingroup$
The problem I find is to show that the quadrilaterals are cyclic from Euclid's books, since it does not seem to be any proposition about it.
$endgroup$
– Smurf
Mar 17 '16 at 7:53
$begingroup$
In this case, it's fairly easy: $A$, $E$, $F$, $C$ all lie on the circle with diameter $AC$, while $B$, $E$, $D$, $F$ lie on the circle with diameter $BD$. In general, you can check whether the two opposite angles add up to $180^circ$ (like in $BEDF$) or the two diagonal-side angles looking at a side are equal (like in $AEFC$.
$endgroup$
– Quang Hoang
Mar 17 '16 at 13:47
$begingroup$
In this case, it's fairly easy: $A$, $E$, $F$, $C$ all lie on the circle with diameter $AC$, while $B$, $E$, $D$, $F$ lie on the circle with diameter $BD$. In general, you can check whether the two opposite angles add up to $180^circ$ (like in $BEDF$) or the two diagonal-side angles looking at a side are equal (like in $AEFC$.
$endgroup$
– Quang Hoang
Mar 17 '16 at 13:47
1
1
$begingroup$
Two things: first I think $angle BDE$ should read $angle BDF$ in line 2.
$endgroup$
– Eric Haney
Mar 20 '16 at 6:18
$begingroup$
Two things: first I think $angle BDE$ should read $angle BDF$ in line 2.
$endgroup$
– Eric Haney
Mar 20 '16 at 6:18
$begingroup$
Also, one way to justify that $AEFC$ is cyclic is to prove the converse of Prop 31 which says the angle in the semicircle is right. Show that a right angle standing on a straight line has its vertex fall on the semicircle described on that straight line as a diameter. In your diagram, we notice that $angle AEC$ and $angle AFC$ are right by construction. This would be sufficient to show that $AEFC$ is a cyclic quadrilateral.
$endgroup$
– Eric Haney
Mar 20 '16 at 6:27
$begingroup$
Also, one way to justify that $AEFC$ is cyclic is to prove the converse of Prop 31 which says the angle in the semicircle is right. Show that a right angle standing on a straight line has its vertex fall on the semicircle described on that straight line as a diameter. In your diagram, we notice that $angle AEC$ and $angle AFC$ are right by construction. This would be sufficient to show that $AEFC$ is a cyclic quadrilateral.
$endgroup$
– Eric Haney
Mar 20 '16 at 6:27
add a comment |
$begingroup$
The easiest way I know of showing the altitudes of $ABC$ are concurrent is (1)Prove the right bisectors of a triangle are concurrent. (2) Draw line $l_A$ thru $A$ parallel to $BC$, line $l_B$ thru $B$ parallel to $CA,$ and $l_C$ thru $C$ parallel to $AB.$ These lines meet pair-wise at points $A',B',C'.$ The altitudes of $ABC$ are the right bisectors of $A'B'C'.$
answered Mar 16 '16 at 22:33
DanielWainfleetDanielWainfleet
35.7k31648
$endgroup$
$begingroup$
There is a lot to argument there. I am trying to do a step by step proof with explicit mention of the propositions used on each step. Also would like to follow the cyclic quadrilaterals path.
$endgroup$
– Smurf
Mar 16 '16 at 22:42
add a comment |
$begingroup$
The easiest way I know of showing the altitudes of $ABC$ are concurrent is (1)Prove the right bisectors of a triangle are concurrent. (2) Draw line $l_A$ thru $A$ parallel to $BC$, line $l_B$ thru $B$ parallel to $CA,$ and $l_C$ thru $C$ parallel to $AB.$ These lines meet pair-wise at points $A',B',C'.$ The altitudes of $ABC$ are the right bisectors of $A'B'C'.$
answered Mar 16 '16 at 22:33
DanielWainfleetDanielWainfleet
35.7k31648
$endgroup$
$begingroup$
There is a lot to argument there. I am trying to do a step by step proof with explicit mention of the propositions used on each step. Also would like to follow the cyclic quadrilaterals path.
$endgroup$
– Smurf
Mar 16 '16 at 22:42
add a comment |
$begingroup$
The easiest way I know of showing the altitudes of $ABC$ are concurrent is (1)Prove the right bisectors of a triangle are concurrent. (2) Draw line $l_A$ thru $A$ parallel to $BC$, line $l_B$ thru $B$ parallel to $CA,$ and $l_C$ thru $C$ parallel to $AB.$ These lines meet pair-wise at points $A',B',C'.$ The altitudes of $ABC$ are the right bisectors of $A'B'C'.$
answered Mar 16 '16 at 22:33
DanielWainfleetDanielWainfleet
35.7k31648
$endgroup$
The easiest way I know of showing the altitudes of $ABC$ are concurrent is (1)Prove the right bisectors of a triangle are concurrent. (2) Draw line $l_A$ thru $A$ parallel to $BC$, line $l_B$ thru $B$ parallel to $CA,$ and $l_C$ thru $C$ parallel to $AB.$ These lines meet pair-wise at points $A',B',C'.$ The altitudes of $ABC$ are the right bisectors of $A'B'C'.$
answered Mar 16 '16 at 22:33
DanielWainfleetDanielWainfleet
35.7k31648
answered Mar 16 '16 at 22:33
DanielWainfleetDanielWainfleet
35.7k31648
answered Mar 16 '16 at 22:33
DanielWainfleetDanielWainfleet
35.7k31648
answered Mar 16 '16 at 22:33
DanielWainfleetDanielWainfleet
35.7k31648
35.7k31648
$begingroup$
There is a lot to argument there. I am trying to do a step by step proof with explicit mention of the propositions used on each step. Also would like to follow the cyclic quadrilaterals path.
$endgroup$
– Smurf
Mar 16 '16 at 22:42
add a comment |
$begingroup$
There is a lot to argument there. I am trying to do a step by step proof with explicit mention of the propositions used on each step. Also would like to follow the cyclic quadrilaterals path.
$endgroup$
– Smurf
Mar 16 '16 at 22:42
$begingroup$
There is a lot to argument there. I am trying to do a step by step proof with explicit mention of the propositions used on each step. Also would like to follow the cyclic quadrilaterals path.
$endgroup$
– Smurf
Mar 16 '16 at 22:42
$begingroup$
There is a lot to argument there. I am trying to do a step by step proof with explicit mention of the propositions used on each step. Also would like to follow the cyclic quadrilaterals path.
$endgroup$
– Smurf
Mar 16 '16 at 22:42
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.
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%2f1700740%2fthree-altitudes-of-a-triangle-are-concurrent%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
$begingroup$
I think you mean AF and GCFD instead of AE and GCED. (The F looks like it could be an E in the diagram.)
$endgroup$
– Théophile
Mar 16 '16 at 22:38
$begingroup$
This doesn't help you, but you could do this just using formulas if you were allowed.
$endgroup$
– barrycarter
Mar 16 '16 at 23:22