Prove if $ABCD$ is a parallelogram, $angle OBA = angle ADO iff angle BOC = angle AOD$
$begingroup$
The following is a lemma someone has presented to me:
Lemma:
Let $ABCD$ be a parallelogram in the plane, and let $O$ be a point outside of the parallelogram.
Prove $angle OBA = angle ADO$ if and only $angle BOC = angle AOD$, where all angles are directed.
For example, here is a diagram: 
However, my attempts to find the proof and name, if one exists, of this lemma have not been successful. I have tried my hand at proving this myself, but I’m not all the best at geometry, so I didn’t make much progress.
Here’s a sketch of what I did:
I considered the isogonal conjugation of the points $A$ and $C$ with respect to triangle $OBD$. If we assume $angle OBA = angle ADO$, then both conjugates lie on the perpendicular bisector of $BC$, and if $angle BOC = angle AOD$, then the isogonal conjugate of $A$, $A^*$, lies on $OC$, and vice versa. However this method falls apart as it doesn’t encode the fact that $AB parallel CD$ in a neat way.
Any hints or input will be appreciated.
geometry euclidean-geometry
edited Dec 27 '18 at 17:48
greedoid
44.2k1155110
asked Dec 27 '18 at 6:07
FigureFigure
205
$endgroup$
|
show 1 more comment
$begingroup$
The following is a lemma someone has presented to me:
Lemma:
Let $ABCD$ be a parallelogram in the plane, and let $O$ be a point outside of the parallelogram.
Prove $angle OBA = angle ADO$ if and only $angle BOC = angle AOD$, where all angles are directed.
For example, here is a diagram:
However, my attempts to find the proof and name, if one exists, of this lemma have not been successful. I have tried my hand at proving this myself, but I’m not all the best at geometry, so I didn’t make much progress.
Here’s a sketch of what I did:
I considered the isogonal conjugation of the points $A$ and $C$ with respect to triangle $OBD$. If we assume $angle OBA = angle ADO$, then both conjugates lie on the perpendicular bisector of $BC$, and if $angle BOC = angle AOD$, then the isogonal conjugate of $A$, $A^*$, lies on $OC$, and vice versa. However this method falls apart as it doesn’t encode the fact that $AB parallel CD$ in a neat way.
Any hints or input will be appreciated.
geometry euclidean-geometry
edited Dec 27 '18 at 17:48
greedoid
44.2k1155110
asked Dec 27 '18 at 6:07
FigureFigure
205
$endgroup$
$begingroup$
Are you sure that the statement is correct? My first drawing shows that it is not.
$endgroup$
– Oldboy
Dec 27 '18 at 6:50
$begingroup$
The statement is definitely not correct. The following could be true: $angle OBA = angle ADO leftrightarrow angle BDC = angle AOD$. That one is trivial to prove.
$endgroup$
– Oldboy
Dec 27 '18 at 7:09
$begingroup$
@Oldboy: Whoops, diagram dependency. I have edited it to a more correct statement, taking directed angles instead. This means that $O$ cannot not lie on the circumcircle of $ABD$, which if $O$ were on, would indeed imply your observation.
$endgroup$
– Figure
Dec 27 '18 at 7:19
$begingroup$
So your parallelogram goes like ABDC, not ABCD.
$endgroup$
– Oldboy
Dec 27 '18 at 7:22
$begingroup$
No, the parallelogram is labelled ABCD. I will try to import an image.
$endgroup$
– Figure
Dec 27 '18 at 7:25
|
show 1 more comment
$begingroup$
The following is a lemma someone has presented to me:
Lemma:
Let $ABCD$ be a parallelogram in the plane, and let $O$ be a point outside of the parallelogram.
Prove $angle OBA = angle ADO$ if and only $angle BOC = angle AOD$, where all angles are directed.
For example, here is a diagram:
However, my attempts to find the proof and name, if one exists, of this lemma have not been successful. I have tried my hand at proving this myself, but I’m not all the best at geometry, so I didn’t make much progress.
Here’s a sketch of what I did:
I considered the isogonal conjugation of the points $A$ and $C$ with respect to triangle $OBD$. If we assume $angle OBA = angle ADO$, then both conjugates lie on the perpendicular bisector of $BC$, and if $angle BOC = angle AOD$, then the isogonal conjugate of $A$, $A^*$, lies on $OC$, and vice versa. However this method falls apart as it doesn’t encode the fact that $AB parallel CD$ in a neat way.
Any hints or input will be appreciated.
geometry euclidean-geometry
edited Dec 27 '18 at 17:48
greedoid
44.2k1155110
asked Dec 27 '18 at 6:07
FigureFigure
205
$endgroup$
The following is a lemma someone has presented to me:
Lemma:
Let $ABCD$ be a parallelogram in the plane, and let $O$ be a point outside of the parallelogram.
Prove $angle OBA = angle ADO$ if and only $angle BOC = angle AOD$, where all angles are directed.
For example, here is a diagram:
However, my attempts to find the proof and name, if one exists, of this lemma have not been successful. I have tried my hand at proving this myself, but I’m not all the best at geometry, so I didn’t make much progress.
Here’s a sketch of what I did:
I considered the isogonal conjugation of the points $A$ and $C$ with respect to triangle $OBD$. If we assume $angle OBA = angle ADO$, then both conjugates lie on the perpendicular bisector of $BC$, and if $angle BOC = angle AOD$, then the isogonal conjugate of $A$, $A^*$, lies on $OC$, and vice versa. However this method falls apart as it doesn’t encode the fact that $AB parallel CD$ in a neat way.
Any hints or input will be appreciated.
geometry euclidean-geometry
geometry euclidean-geometry
edited Dec 27 '18 at 17:48
greedoid
44.2k1155110
asked Dec 27 '18 at 6:07
FigureFigure
205
edited Dec 27 '18 at 17:48
greedoid
44.2k1155110
asked Dec 27 '18 at 6:07
FigureFigure
205
edited Dec 27 '18 at 17:48
greedoid
44.2k1155110
edited Dec 27 '18 at 17:48
greedoid
44.2k1155110
edited Dec 27 '18 at 17:48
greedoid
44.2k1155110
44.2k1155110
asked Dec 27 '18 at 6:07
FigureFigure
205
asked Dec 27 '18 at 6:07
FigureFigure
205
asked Dec 27 '18 at 6:07
FigureFigure
205
205
$begingroup$
Are you sure that the statement is correct? My first drawing shows that it is not.
$endgroup$
– Oldboy
Dec 27 '18 at 6:50
$begingroup$
The statement is definitely not correct. The following could be true: $angle OBA = angle ADO leftrightarrow angle BDC = angle AOD$. That one is trivial to prove.
$endgroup$
– Oldboy
Dec 27 '18 at 7:09
$begingroup$
@Oldboy: Whoops, diagram dependency. I have edited it to a more correct statement, taking directed angles instead. This means that $O$ cannot not lie on the circumcircle of $ABD$, which if $O$ were on, would indeed imply your observation.
$endgroup$
– Figure
Dec 27 '18 at 7:19
$begingroup$
So your parallelogram goes like ABDC, not ABCD.
$endgroup$
– Oldboy
Dec 27 '18 at 7:22
$begingroup$
No, the parallelogram is labelled ABCD. I will try to import an image.
$endgroup$
– Figure
Dec 27 '18 at 7:25
|
show 1 more comment
$begingroup$
Are you sure that the statement is correct? My first drawing shows that it is not.
$endgroup$
– Oldboy
Dec 27 '18 at 6:50
$begingroup$
The statement is definitely not correct. The following could be true: $angle OBA = angle ADO leftrightarrow angle BDC = angle AOD$. That one is trivial to prove.
$endgroup$
– Oldboy
Dec 27 '18 at 7:09
$begingroup$
@Oldboy: Whoops, diagram dependency. I have edited it to a more correct statement, taking directed angles instead. This means that $O$ cannot not lie on the circumcircle of $ABD$, which if $O$ were on, would indeed imply your observation.
$endgroup$
– Figure
Dec 27 '18 at 7:19
$begingroup$
So your parallelogram goes like ABDC, not ABCD.
$endgroup$
– Oldboy
Dec 27 '18 at 7:22
$begingroup$
No, the parallelogram is labelled ABCD. I will try to import an image.
$endgroup$
– Figure
Dec 27 '18 at 7:25
$begingroup$
Are you sure that the statement is correct? My first drawing shows that it is not.
$endgroup$
– Oldboy
Dec 27 '18 at 6:50
$begingroup$
Are you sure that the statement is correct? My first drawing shows that it is not.
$endgroup$
– Oldboy
Dec 27 '18 at 6:50
$begingroup$
The statement is definitely not correct. The following could be true: $angle OBA = angle ADO leftrightarrow angle BDC = angle AOD$. That one is trivial to prove.
$endgroup$
– Oldboy
Dec 27 '18 at 7:09
$begingroup$
The statement is definitely not correct. The following could be true: $angle OBA = angle ADO leftrightarrow angle BDC = angle AOD$. That one is trivial to prove.
$endgroup$
– Oldboy
Dec 27 '18 at 7:09
$begingroup$
@Oldboy: Whoops, diagram dependency. I have edited it to a more correct statement, taking directed angles instead. This means that $O$ cannot not lie on the circumcircle of $ABD$, which if $O$ were on, would indeed imply your observation.
$endgroup$
– Figure
Dec 27 '18 at 7:19
$begingroup$
@Oldboy: Whoops, diagram dependency. I have edited it to a more correct statement, taking directed angles instead. This means that $O$ cannot not lie on the circumcircle of $ABD$, which if $O$ were on, would indeed imply your observation.
$endgroup$
– Figure
Dec 27 '18 at 7:19
$begingroup$
So your parallelogram goes like ABDC, not ABCD.
$endgroup$
– Oldboy
Dec 27 '18 at 7:22
$begingroup$
So your parallelogram goes like ABDC, not ABCD.
$endgroup$
– Oldboy
Dec 27 '18 at 7:22
$begingroup$
No, the parallelogram is labelled ABCD. I will try to import an image.
$endgroup$
– Figure
Dec 27 '18 at 7:25
$begingroup$
No, the parallelogram is labelled ABCD. I will try to import an image.
$endgroup$
– Figure
Dec 27 '18 at 7:25
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
Let $X$ be the midpoint of $AC$. Let $A', C'$ be the reflections of $A$ and $C$ with respect to $OD, OB$, respectively. Let $Y, Z$ be the midpoints of $CC', AA'$, respectively. Note that $Y, Z$ are the perpendicular projections of $C, A$ onto $OB, OD$, repsectively.
You can prove that $angle BOC = angle AOD iff triangle AOC' equiv triangle A'OC iff AC' = A'C iff XY=XZ.$
Now, construct a parallelogram $AOCO'$ and note that $angle ODA = angle O'BC$. This is equivalent to $angle OBC = angle ABO'$. Let $T$ be the projection of $A$ onto $O'B$. Repeat the argument above to show the equivalence $angle OBC = angle ABO' iff XY = XT$. Combine all ingredients keeping in mind that $XT=XZ$ due to symmetry.
answered Dec 27 '18 at 8:42
timon92timon92
4,3981826
$endgroup$
add a comment |
$begingroup$
Translate $O$ for vector $AB$ to $O'$. Then $angle BCO' = angle ADO$ and $angle BO'C = angle AOD$.
Then $angle AOD = angle BOC iff angle BO'C = angle BOC$ and that is iff $OO'BC$ is cyclic and that is iff $angle BCO' = angle BOO' = angle OBA$ (since $AB||OO'$).
This is the end of a prove.
answered Dec 27 '18 at 14:30
greedoidgreedoid
44.2k1155110
$endgroup$
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%2f3053643%2fprove-if-abcd-is-a-parallelogram-angle-oba-angle-ado-iff-angle-boc%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
$begingroup$
Let $X$ be the midpoint of $AC$. Let $A', C'$ be the reflections of $A$ and $C$ with respect to $OD, OB$, respectively. Let $Y, Z$ be the midpoints of $CC', AA'$, respectively. Note that $Y, Z$ are the perpendicular projections of $C, A$ onto $OB, OD$, repsectively.
You can prove that $angle BOC = angle AOD iff triangle AOC' equiv triangle A'OC iff AC' = A'C iff XY=XZ.$
Now, construct a parallelogram $AOCO'$ and note that $angle ODA = angle O'BC$. This is equivalent to $angle OBC = angle ABO'$. Let $T$ be the projection of $A$ onto $O'B$. Repeat the argument above to show the equivalence $angle OBC = angle ABO' iff XY = XT$. Combine all ingredients keeping in mind that $XT=XZ$ due to symmetry.
answered Dec 27 '18 at 8:42
timon92timon92
4,3981826
$endgroup$
add a comment |
$begingroup$
Let $X$ be the midpoint of $AC$. Let $A', C'$ be the reflections of $A$ and $C$ with respect to $OD, OB$, respectively. Let $Y, Z$ be the midpoints of $CC', AA'$, respectively. Note that $Y, Z$ are the perpendicular projections of $C, A$ onto $OB, OD$, repsectively.
You can prove that $angle BOC = angle AOD iff triangle AOC' equiv triangle A'OC iff AC' = A'C iff XY=XZ.$
Now, construct a parallelogram $AOCO'$ and note that $angle ODA = angle O'BC$. This is equivalent to $angle OBC = angle ABO'$. Let $T$ be the projection of $A$ onto $O'B$. Repeat the argument above to show the equivalence $angle OBC = angle ABO' iff XY = XT$. Combine all ingredients keeping in mind that $XT=XZ$ due to symmetry.
answered Dec 27 '18 at 8:42
timon92timon92
4,3981826
$endgroup$
add a comment |
$begingroup$
Let $X$ be the midpoint of $AC$. Let $A', C'$ be the reflections of $A$ and $C$ with respect to $OD, OB$, respectively. Let $Y, Z$ be the midpoints of $CC', AA'$, respectively. Note that $Y, Z$ are the perpendicular projections of $C, A$ onto $OB, OD$, repsectively.
You can prove that $angle BOC = angle AOD iff triangle AOC' equiv triangle A'OC iff AC' = A'C iff XY=XZ.$
Now, construct a parallelogram $AOCO'$ and note that $angle ODA = angle O'BC$. This is equivalent to $angle OBC = angle ABO'$. Let $T$ be the projection of $A$ onto $O'B$. Repeat the argument above to show the equivalence $angle OBC = angle ABO' iff XY = XT$. Combine all ingredients keeping in mind that $XT=XZ$ due to symmetry.
answered Dec 27 '18 at 8:42
timon92timon92
4,3981826
$endgroup$
Let $X$ be the midpoint of $AC$. Let $A', C'$ be the reflections of $A$ and $C$ with respect to $OD, OB$, respectively. Let $Y, Z$ be the midpoints of $CC', AA'$, respectively. Note that $Y, Z$ are the perpendicular projections of $C, A$ onto $OB, OD$, repsectively.
You can prove that $angle BOC = angle AOD iff triangle AOC' equiv triangle A'OC iff AC' = A'C iff XY=XZ.$
Now, construct a parallelogram $AOCO'$ and note that $angle ODA = angle O'BC$. This is equivalent to $angle OBC = angle ABO'$. Let $T$ be the projection of $A$ onto $O'B$. Repeat the argument above to show the equivalence $angle OBC = angle ABO' iff XY = XT$. Combine all ingredients keeping in mind that $XT=XZ$ due to symmetry.
answered Dec 27 '18 at 8:42
timon92timon92
4,3981826
edited Dec 27 '18 at 10:04
answered Dec 27 '18 at 8:42
timon92timon92
4,3981826
answered Dec 27 '18 at 8:42
timon92timon92
4,3981826
answered Dec 27 '18 at 8:42
timon92timon92
4,3981826
4,3981826
add a comment |
add a comment |
$begingroup$
Translate $O$ for vector $AB$ to $O'$. Then $angle BCO' = angle ADO$ and $angle BO'C = angle AOD$.
Then $angle AOD = angle BOC iff angle BO'C = angle BOC$ and that is iff $OO'BC$ is cyclic and that is iff $angle BCO' = angle BOO' = angle OBA$ (since $AB||OO'$).
This is the end of a prove.
answered Dec 27 '18 at 14:30
greedoidgreedoid
44.2k1155110
$endgroup$
add a comment |
$begingroup$
Translate $O$ for vector $AB$ to $O'$. Then $angle BCO' = angle ADO$ and $angle BO'C = angle AOD$.
Then $angle AOD = angle BOC iff angle BO'C = angle BOC$ and that is iff $OO'BC$ is cyclic and that is iff $angle BCO' = angle BOO' = angle OBA$ (since $AB||OO'$).
This is the end of a prove.
answered Dec 27 '18 at 14:30
greedoidgreedoid
44.2k1155110
$endgroup$
add a comment |
$begingroup$
Translate $O$ for vector $AB$ to $O'$. Then $angle BCO' = angle ADO$ and $angle BO'C = angle AOD$.
Then $angle AOD = angle BOC iff angle BO'C = angle BOC$ and that is iff $OO'BC$ is cyclic and that is iff $angle BCO' = angle BOO' = angle OBA$ (since $AB||OO'$).
This is the end of a prove.
answered Dec 27 '18 at 14:30
greedoidgreedoid
44.2k1155110
$endgroup$
Translate $O$ for vector $AB$ to $O'$. Then $angle BCO' = angle ADO$ and $angle BO'C = angle AOD$.
Then $angle AOD = angle BOC iff angle BO'C = angle BOC$ and that is iff $OO'BC$ is cyclic and that is iff $angle BCO' = angle BOO' = angle OBA$ (since $AB||OO'$).
This is the end of a prove.
answered Dec 27 '18 at 14:30
greedoidgreedoid
44.2k1155110
edited Dec 29 '18 at 10:04
answered Dec 27 '18 at 14:30
greedoidgreedoid
44.2k1155110
answered Dec 27 '18 at 14:30
greedoidgreedoid
44.2k1155110
answered Dec 27 '18 at 14:30
greedoidgreedoid
44.2k1155110
44.2k1155110
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.
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%2f3053643%2fprove-if-abcd-is-a-parallelogram-angle-oba-angle-ado-iff-angle-boc%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
$begingroup$
Are you sure that the statement is correct? My first drawing shows that it is not.
$endgroup$
– Oldboy
Dec 27 '18 at 6:50
$begingroup$
The statement is definitely not correct. The following could be true: $angle OBA = angle ADO leftrightarrow angle BDC = angle AOD$. That one is trivial to prove.
$endgroup$
– Oldboy
Dec 27 '18 at 7:09
$begingroup$
@Oldboy: Whoops, diagram dependency. I have edited it to a more correct statement, taking directed angles instead. This means that $O$ cannot not lie on the circumcircle of $ABD$, which if $O$ were on, would indeed imply your observation.
$endgroup$
– Figure
Dec 27 '18 at 7:19
$begingroup$
So your parallelogram goes like ABDC, not ABCD.
$endgroup$
– Oldboy
Dec 27 '18 at 7:22
$begingroup$
No, the parallelogram is labelled ABCD. I will try to import an image.
$endgroup$
– Figure
Dec 27 '18 at 7:25