Prove if $ABCD$ is a parallelogram, $angle OBA = angle ADO iff angle BOC = angle AOD$
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
38.2k114795
asked Dec 27 '18 at 6:07
Figure
205
|
show 1 more comment
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
38.2k114795
asked Dec 27 '18 at 6:07
Figure
205
Are you sure that the statement is correct? My first drawing shows that it is not.
– Oldboy
Dec 27 '18 at 6:50
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.
– Oldboy
Dec 27 '18 at 7:09
@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.
– Figure
Dec 27 '18 at 7:19
So your parallelogram goes like ABDC, not ABCD.
– Oldboy
Dec 27 '18 at 7:22
No, the parallelogram is labelled ABCD. I will try to import an image.
– Figure
Dec 27 '18 at 7:25
|
show 1 more comment
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
38.2k114795
asked Dec 27 '18 at 6:07
Figure
205
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
38.2k114795
asked Dec 27 '18 at 6:07
Figure
205
edited Dec 27 '18 at 17:48
greedoid
38.2k114795
asked Dec 27 '18 at 6:07
Figure
205
edited Dec 27 '18 at 17:48
greedoid
38.2k114795
edited Dec 27 '18 at 17:48
greedoid
38.2k114795
edited Dec 27 '18 at 17:48
greedoid
38.2k114795
38.2k114795
asked Dec 27 '18 at 6:07
Figure
205
asked Dec 27 '18 at 6:07
Figure
205
asked Dec 27 '18 at 6:07
Figure
205
205
Are you sure that the statement is correct? My first drawing shows that it is not.
– Oldboy
Dec 27 '18 at 6:50
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.
– Oldboy
Dec 27 '18 at 7:09
@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.
– Figure
Dec 27 '18 at 7:19
So your parallelogram goes like ABDC, not ABCD.
– Oldboy
Dec 27 '18 at 7:22
No, the parallelogram is labelled ABCD. I will try to import an image.
– Figure
Dec 27 '18 at 7:25
|
show 1 more comment
Are you sure that the statement is correct? My first drawing shows that it is not.
– Oldboy
Dec 27 '18 at 6:50
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.
– Oldboy
Dec 27 '18 at 7:09
@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.
– Figure
Dec 27 '18 at 7:19
So your parallelogram goes like ABDC, not ABCD.
– Oldboy
Dec 27 '18 at 7:22
No, the parallelogram is labelled ABCD. I will try to import an image.
– Figure
Dec 27 '18 at 7:25
Are you sure that the statement is correct? My first drawing shows that it is not.
– Oldboy
Dec 27 '18 at 6:50
Are you sure that the statement is correct? My first drawing shows that it is not.
– Oldboy
Dec 27 '18 at 6:50
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.
– Oldboy
Dec 27 '18 at 7:09
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.
– Oldboy
Dec 27 '18 at 7:09
@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.
– Figure
Dec 27 '18 at 7:19
@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.
– Figure
Dec 27 '18 at 7:19
So your parallelogram goes like ABDC, not ABCD.
– Oldboy
Dec 27 '18 at 7:22
So your parallelogram goes like ABDC, not ABCD.
– Oldboy
Dec 27 '18 at 7:22
No, the parallelogram is labelled ABCD. I will try to import an image.
– Figure
Dec 27 '18 at 7:25
No, the parallelogram is labelled ABCD. I will try to import an image.
– Figure
Dec 27 '18 at 7:25
|
show 1 more comment
2 Answers
2
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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
timon92
4,3181726
add a comment |
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
greedoid
38.2k114795
add a comment |
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2 Answers
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2 Answers
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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
timon92
4,3181726
add a comment |
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
timon92
4,3181726
add a comment |
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
timon92
4,3181726
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
timon92
4,3181726
edited Dec 27 '18 at 10:04
answered Dec 27 '18 at 8:42
timon92
4,3181726
answered Dec 27 '18 at 8:42
timon92
4,3181726
answered Dec 27 '18 at 8:42
timon92
4,3181726
4,3181726
add a comment |
add a comment |
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
greedoid
38.2k114795
add a comment |
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
greedoid
38.2k114795
add a comment |
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
greedoid
38.2k114795
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
greedoid
38.2k114795
edited Dec 29 '18 at 10:04
answered Dec 27 '18 at 14:30
greedoid
38.2k114795
answered Dec 27 '18 at 14:30
greedoid
38.2k114795
answered Dec 27 '18 at 14:30
greedoid
38.2k114795
38.2k114795
add a comment |
add a comment |
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Are you sure that the statement is correct? My first drawing shows that it is not.
– Oldboy
Dec 27 '18 at 6:50
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.
– Oldboy
Dec 27 '18 at 7:09
@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.
– Figure
Dec 27 '18 at 7:19
So your parallelogram goes like ABDC, not ABCD.
– Oldboy
Dec 27 '18 at 7:22
No, the parallelogram is labelled ABCD. I will try to import an image.
– Figure
Dec 27 '18 at 7:25