Proving colinearity of 3 points(basic Euclidean geometry)
1
$begingroup$
In the above image, AP is bisector of the angle BAC and DP is a perpendicular bisector of the segment BC.
How can it be proved that the points E, D, F are colinear?
euclidean-geometry
asked Dec 3 '18 at 10:02
Ki Yoon EumKi Yoon Eum
175
$endgroup$
add a comment |
1
$begingroup$
In the above image, AP is bisector of the angle BAC and DP is a perpendicular bisector of the segment BC.
How can it be proved that the points E, D, F are colinear?
euclidean-geometry
asked Dec 3 '18 at 10:02
Ki Yoon EumKi Yoon Eum
175
$endgroup$
1
$begingroup$
You cannot in general ! because it is true only for some special position of $BC$
$endgroup$
– Emilio Novati
Dec 3 '18 at 10:29
$begingroup$
What special position? This is an exercise problem in "Euclidean and non-Euclidean geometry" and there is no such mention
$endgroup$
– Ki Yoon Eum
Dec 3 '18 at 11:23
1
$begingroup$
And... I solve it actually using Menelaus' theorem!
$endgroup$
– Ki Yoon Eum
Dec 3 '18 at 12:23
$begingroup$
Can you add your solution (as an answer)?
$endgroup$
– Berci
Dec 3 '18 at 19:32
add a comment |
1
1
1
$begingroup$
In the above image, AP is bisector of the angle BAC and DP is a perpendicular bisector of the segment BC.
How can it be proved that the points E, D, F are colinear?
euclidean-geometry
asked Dec 3 '18 at 10:02
Ki Yoon EumKi Yoon Eum
175
$endgroup$
In the above image, AP is bisector of the angle BAC and DP is a perpendicular bisector of the segment BC.
How can it be proved that the points E, D, F are colinear?
euclidean-geometry
euclidean-geometry
asked Dec 3 '18 at 10:02
Ki Yoon EumKi Yoon Eum
175
asked Dec 3 '18 at 10:02
Ki Yoon EumKi Yoon Eum
175
asked Dec 3 '18 at 10:02
Ki Yoon EumKi Yoon Eum
175
asked Dec 3 '18 at 10:02
Ki Yoon EumKi Yoon Eum
175
asked Dec 3 '18 at 10:02
Ki Yoon EumKi Yoon Eum
175
175
1
$begingroup$
You cannot in general ! because it is true only for some special position of $BC$
$endgroup$
– Emilio Novati
Dec 3 '18 at 10:29
$begingroup$
What special position? This is an exercise problem in "Euclidean and non-Euclidean geometry" and there is no such mention
$endgroup$
– Ki Yoon Eum
Dec 3 '18 at 11:23
1
$begingroup$
And... I solve it actually using Menelaus' theorem!
$endgroup$
– Ki Yoon Eum
Dec 3 '18 at 12:23
$begingroup$
Can you add your solution (as an answer)?
$endgroup$
– Berci
Dec 3 '18 at 19:32
add a comment |
1
$begingroup$
You cannot in general ! because it is true only for some special position of $BC$
$endgroup$
– Emilio Novati
Dec 3 '18 at 10:29
$begingroup$
What special position? This is an exercise problem in "Euclidean and non-Euclidean geometry" and there is no such mention
$endgroup$
– Ki Yoon Eum
Dec 3 '18 at 11:23
1
$begingroup$
And... I solve it actually using Menelaus' theorem!
$endgroup$
– Ki Yoon Eum
Dec 3 '18 at 12:23
$begingroup$
Can you add your solution (as an answer)?
$endgroup$
– Berci
Dec 3 '18 at 19:32
1
1
$begingroup$
You cannot in general ! because it is true only for some special position of $BC$
$endgroup$
– Emilio Novati
Dec 3 '18 at 10:29
$begingroup$
You cannot in general ! because it is true only for some special position of $BC$
$endgroup$
– Emilio Novati
Dec 3 '18 at 10:29
$begingroup$
What special position? This is an exercise problem in "Euclidean and non-Euclidean geometry" and there is no such mention
$endgroup$
– Ki Yoon Eum
Dec 3 '18 at 11:23
$begingroup$
What special position? This is an exercise problem in "Euclidean and non-Euclidean geometry" and there is no such mention
$endgroup$
– Ki Yoon Eum
Dec 3 '18 at 11:23
1
1
$begingroup$
And... I solve it actually using Menelaus' theorem!
$endgroup$
– Ki Yoon Eum
Dec 3 '18 at 12:23
$begingroup$
And... I solve it actually using Menelaus' theorem!
$endgroup$
– Ki Yoon Eum
Dec 3 '18 at 12:23
$begingroup$
Can you add your solution (as an answer)?
$endgroup$
– Berci
Dec 3 '18 at 19:32
$begingroup$
Can you add your solution (as an answer)?
$endgroup$
– Berci
Dec 3 '18 at 19:32
add a comment |
1 Answer
1
active
oldest
votes
0
$begingroup$
Note that $P$ is the mid-point of the minor arc BC of the circumcircle ABC. So EDF is the Simson line of the point $P$.
answered Dec 4 '18 at 10:52
user10354138user10354138
7,4322925
$endgroup$
$begingroup$
So, how is the Simpson "line" a line?
$endgroup$
– Oscar Lanzi
Dec 4 '18 at 13:55
$begingroup$
The linked wiki article has a proof, as does many other sites such as cut-the-knot. I don't think I should regurgitate it here.
$endgroup$
– user10354138
Dec 4 '18 at 13:57
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
});
}
});
draft saved
draft discarded
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%2f3023864%2fproving-colinearity-of-3-pointsbasic-euclidean-geometry%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
0
$begingroup$
Note that $P$ is the mid-point of the minor arc BC of the circumcircle ABC. So EDF is the Simson line of the point $P$.
answered Dec 4 '18 at 10:52
user10354138user10354138
7,4322925
$endgroup$
$begingroup$
So, how is the Simpson "line" a line?
$endgroup$
– Oscar Lanzi
Dec 4 '18 at 13:55
$begingroup$
The linked wiki article has a proof, as does many other sites such as cut-the-knot. I don't think I should regurgitate it here.
$endgroup$
– user10354138
Dec 4 '18 at 13:57
add a comment |
0
$begingroup$
Note that $P$ is the mid-point of the minor arc BC of the circumcircle ABC. So EDF is the Simson line of the point $P$.
answered Dec 4 '18 at 10:52
user10354138user10354138
7,4322925
$endgroup$
$begingroup$
So, how is the Simpson "line" a line?
$endgroup$
– Oscar Lanzi
Dec 4 '18 at 13:55
$begingroup$
The linked wiki article has a proof, as does many other sites such as cut-the-knot. I don't think I should regurgitate it here.
$endgroup$
– user10354138
Dec 4 '18 at 13:57
add a comment |
0
0
0
$begingroup$
Note that $P$ is the mid-point of the minor arc BC of the circumcircle ABC. So EDF is the Simson line of the point $P$.
answered Dec 4 '18 at 10:52
user10354138user10354138
7,4322925
$endgroup$
Note that $P$ is the mid-point of the minor arc BC of the circumcircle ABC. So EDF is the Simson line of the point $P$.
answered Dec 4 '18 at 10:52
user10354138user10354138
7,4322925
answered Dec 4 '18 at 10:52
user10354138user10354138
7,4322925
answered Dec 4 '18 at 10:52
user10354138user10354138
7,4322925
answered Dec 4 '18 at 10:52
user10354138user10354138
7,4322925
7,4322925
$begingroup$
So, how is the Simpson "line" a line?
$endgroup$
– Oscar Lanzi
Dec 4 '18 at 13:55
$begingroup$
The linked wiki article has a proof, as does many other sites such as cut-the-knot. I don't think I should regurgitate it here.
$endgroup$
– user10354138
Dec 4 '18 at 13:57
add a comment |
$begingroup$
So, how is the Simpson "line" a line?
$endgroup$
– Oscar Lanzi
Dec 4 '18 at 13:55
$begingroup$
The linked wiki article has a proof, as does many other sites such as cut-the-knot. I don't think I should regurgitate it here.
$endgroup$
– user10354138
Dec 4 '18 at 13:57
$begingroup$
So, how is the Simpson "line" a line?
$endgroup$
– Oscar Lanzi
Dec 4 '18 at 13:55
$begingroup$
So, how is the Simpson "line" a line?
$endgroup$
– Oscar Lanzi
Dec 4 '18 at 13:55
$begingroup$
The linked wiki article has a proof, as does many other sites such as cut-the-knot. I don't think I should regurgitate it here.
$endgroup$
– user10354138
Dec 4 '18 at 13:57
$begingroup$
The linked wiki article has a proof, as does many other sites such as cut-the-knot. I don't think I should regurgitate it here.
$endgroup$
– user10354138
Dec 4 '18 at 13:57
add a comment |
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
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%2f3023864%2fproving-colinearity-of-3-pointsbasic-euclidean-geometry%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
$begingroup$
You cannot in general ! because it is true only for some special position of $BC$
$endgroup$
– Emilio Novati
Dec 3 '18 at 10:29
$begingroup$
What special position? This is an exercise problem in "Euclidean and non-Euclidean geometry" and there is no such mention
$endgroup$
– Ki Yoon Eum
Dec 3 '18 at 11:23
1
$begingroup$
And... I solve it actually using Menelaus' theorem!
$endgroup$
– Ki Yoon Eum
Dec 3 '18 at 12:23
$begingroup$
Can you add your solution (as an answer)?
$endgroup$
– Berci
Dec 3 '18 at 19:32