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Construct a quadrilateral, not a parallelogram, in which pair of opposite angles and a pair of opposite sides...
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This question already has an answer here:
If a quadrilateral has a pair of equal opposite sides, and a pair of equal opposite angles, then is it necessarily a parallelogram?
3 answers
I want to construct a quadrilateral which is not a parallelogram, in which a pair of opposite angles and a pair of opposite sides are equal. I tried drawing one, but I am not able to. Please help. (And I dont think I will be able to post what I have tried or drawn. Hope the statement of the post is clear).
Umm, I understand that it might look like it is the duplicate of another similar question( I saw that post actually!). But I do not think it explained how to construct one. I know such a quadrilateral exists, that is not my real question. I want to know how it is constructed. I hope I made it clear.
Also, I wanted a short, simplified and clear solution for this(I can elaborate that).
geometry euclidean-geometry examples-counterexamples quadrilateral
asked Jan 8 at 12:32
YellowYellow
16011
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marked as duplicate by Saad, Aretino, Eevee Trainer, Adrian Keister, José Carlos Santos Jan 9 at 17:53
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
If a quadrilateral has a pair of equal opposite sides, and a pair of equal opposite angles, then is it necessarily a parallelogram?
3 answers
I want to construct a quadrilateral which is not a parallelogram, in which a pair of opposite angles and a pair of opposite sides are equal. I tried drawing one, but I am not able to. Please help. (And I dont think I will be able to post what I have tried or drawn. Hope the statement of the post is clear).
Umm, I understand that it might look like it is the duplicate of another similar question( I saw that post actually!). But I do not think it explained how to construct one. I know such a quadrilateral exists, that is not my real question. I want to know how it is constructed. I hope I made it clear.
Also, I wanted a short, simplified and clear solution for this(I can elaborate that).
geometry euclidean-geometry examples-counterexamples quadrilateral
asked Jan 8 at 12:32
YellowYellow
16011
$endgroup$
marked as duplicate by Saad, Aretino, Eevee Trainer, Adrian Keister, José Carlos Santos Jan 9 at 17:53
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
Related/Duplicate: "If a quadrilateral has a pair of equal opposite sides, and a pair of equal opposite angles, then is it necessarily a parallelogram?" and "Construct or prove existence of a certain quadrilateral". Perhaps others.
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– Blue
Jan 8 at 13:28
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Is it fine now?
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– Yellow
Jan 8 at 15:24
add a comment |
$begingroup$
This question already has an answer here:
If a quadrilateral has a pair of equal opposite sides, and a pair of equal opposite angles, then is it necessarily a parallelogram?
3 answers
I want to construct a quadrilateral which is not a parallelogram, in which a pair of opposite angles and a pair of opposite sides are equal. I tried drawing one, but I am not able to. Please help. (And I dont think I will be able to post what I have tried or drawn. Hope the statement of the post is clear).
Umm, I understand that it might look like it is the duplicate of another similar question( I saw that post actually!). But I do not think it explained how to construct one. I know such a quadrilateral exists, that is not my real question. I want to know how it is constructed. I hope I made it clear.
Also, I wanted a short, simplified and clear solution for this(I can elaborate that).
geometry euclidean-geometry examples-counterexamples quadrilateral
asked Jan 8 at 12:32
YellowYellow
16011
$endgroup$
This question already has an answer here:
If a quadrilateral has a pair of equal opposite sides, and a pair of equal opposite angles, then is it necessarily a parallelogram?
3 answers
I want to construct a quadrilateral which is not a parallelogram, in which a pair of opposite angles and a pair of opposite sides are equal. I tried drawing one, but I am not able to. Please help. (And I dont think I will be able to post what I have tried or drawn. Hope the statement of the post is clear).
Umm, I understand that it might look like it is the duplicate of another similar question( I saw that post actually!). But I do not think it explained how to construct one. I know such a quadrilateral exists, that is not my real question. I want to know how it is constructed. I hope I made it clear.
Also, I wanted a short, simplified and clear solution for this(I can elaborate that).
This question already has an answer here:
If a quadrilateral has a pair of equal opposite sides, and a pair of equal opposite angles, then is it necessarily a parallelogram?
3 answers
geometry euclidean-geometry examples-counterexamples quadrilateral
geometry euclidean-geometry examples-counterexamples quadrilateral
asked Jan 8 at 12:32
YellowYellow
16011
asked Jan 8 at 12:32
YellowYellow
16011
edited Jan 8 at 15:53
Yellow
asked Jan 8 at 12:32
YellowYellow
16011
asked Jan 8 at 12:32
YellowYellow
16011
asked Jan 8 at 12:32
YellowYellow
16011
16011
marked as duplicate by Saad, Aretino, Eevee Trainer, Adrian Keister, José Carlos Santos Jan 9 at 17:53
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Saad, Aretino, Eevee Trainer, Adrian Keister, José Carlos Santos Jan 9 at 17:53
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
Related/Duplicate: "If a quadrilateral has a pair of equal opposite sides, and a pair of equal opposite angles, then is it necessarily a parallelogram?" and "Construct or prove existence of a certain quadrilateral". Perhaps others.
$endgroup$
– Blue
Jan 8 at 13:28
$begingroup$
Is it fine now?
$endgroup$
– Yellow
Jan 8 at 15:24
add a comment |
$begingroup$
Related/Duplicate: "If a quadrilateral has a pair of equal opposite sides, and a pair of equal opposite angles, then is it necessarily a parallelogram?" and "Construct or prove existence of a certain quadrilateral". Perhaps others.
$endgroup$
– Blue
Jan 8 at 13:28
$begingroup$
Is it fine now?
$endgroup$
– Yellow
Jan 8 at 15:24
$begingroup$
Related/Duplicate: "If a quadrilateral has a pair of equal opposite sides, and a pair of equal opposite angles, then is it necessarily a parallelogram?" and "Construct or prove existence of a certain quadrilateral". Perhaps others.
$endgroup$
– Blue
Jan 8 at 13:28
$begingroup$
Related/Duplicate: "If a quadrilateral has a pair of equal opposite sides, and a pair of equal opposite angles, then is it necessarily a parallelogram?" and "Construct or prove existence of a certain quadrilateral". Perhaps others.
$endgroup$
– Blue
Jan 8 at 13:28
$begingroup$
Is it fine now?
$endgroup$
– Yellow
Jan 8 at 15:24
$begingroup$
Is it fine now?
$endgroup$
– Yellow
Jan 8 at 15:24
add a comment |
2 Answers
2
active
oldest
votes
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Take $Delta ABK$ such that $measuredangle ABK>90^{circ}$ and let $Din AK$ such that $BD=BK$.
Now, let $Delta BDCcongDelta KBA$ such that $K$ and $C$ are placed in the same side respect to the line $DB$.
Thus, $ABCD$ is a needed quadrilateral because $AB=DC$ and $measuredangle A=measuredangle C$.
Easy to see that $ABCD$ is not parallelogram.
answered Jan 8 at 12:54
Michael RozenbergMichael Rozenberg
100k1591192
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$ABCD$ will always turn out to be concave?( I mean, I get a concave quadrilateral, and the nomenclature also differs, thats why)
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– Yellow
Jan 8 at 15:40
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And also, can you explain how $angle A$ = $angle C$?
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– Yellow
Jan 8 at 15:41
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@Anu Radha Yes, it concave. The equality of angles follows from congruent triangles.
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– Michael Rozenberg
Jan 8 at 16:29
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Is there any particular reason why the resulting quadrilateral is concave rather than being convex? Why can it NOT be convex?
$endgroup$
– Yellow
Jan 8 at 17:09
$begingroup$
Also, by $angle A$ and $angle C$, which two angles do you mean? I dont understand how equality follows from the congruency..(Sorry, really!)
$endgroup$
– Yellow
Jan 8 at 18:39
|
show 2 more comments
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You are not succeeding in finding such a quadrilateral because none exists.
If $ABCD$ is such a quadrilateral with $|AD| = |BC|$ and $beta = delta$ then the triangles $ABC$ and $CDA$ are congruent. It follows that $sphericalangle CAB = sphericalangle DCA$, hence $AB$ is parallel to $DC$ from this you can easily deduce that the quadrilateral is a parallelogram.
answered Jan 8 at 12:53
Sergey YurkevichSergey Yurkevich
162
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They are not always congruent. See please my counterexample.
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– Michael Rozenberg
Jan 8 at 13:06
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Take $Delta ABK$ such that $measuredangle ABK>90^{circ}$ and let $Din AK$ such that $BD=BK$.
Now, let $Delta BDCcongDelta KBA$ such that $K$ and $C$ are placed in the same side respect to the line $DB$.
Thus, $ABCD$ is a needed quadrilateral because $AB=DC$ and $measuredangle A=measuredangle C$.
Easy to see that $ABCD$ is not parallelogram.
answered Jan 8 at 12:54
Michael RozenbergMichael Rozenberg
100k1591192
$endgroup$
$begingroup$
$ABCD$ will always turn out to be concave?( I mean, I get a concave quadrilateral, and the nomenclature also differs, thats why)
$endgroup$
– Yellow
Jan 8 at 15:40
$begingroup$
And also, can you explain how $angle A$ = $angle C$?
$endgroup$
– Yellow
Jan 8 at 15:41
$begingroup$
@Anu Radha Yes, it concave. The equality of angles follows from congruent triangles.
$endgroup$
– Michael Rozenberg
Jan 8 at 16:29
$begingroup$
Is there any particular reason why the resulting quadrilateral is concave rather than being convex? Why can it NOT be convex?
$endgroup$
– Yellow
Jan 8 at 17:09
$begingroup$
Also, by $angle A$ and $angle C$, which two angles do you mean? I dont understand how equality follows from the congruency..(Sorry, really!)
$endgroup$
– Yellow
Jan 8 at 18:39
|
show 2 more comments
$begingroup$
Take $Delta ABK$ such that $measuredangle ABK>90^{circ}$ and let $Din AK$ such that $BD=BK$.
Now, let $Delta BDCcongDelta KBA$ such that $K$ and $C$ are placed in the same side respect to the line $DB$.
Thus, $ABCD$ is a needed quadrilateral because $AB=DC$ and $measuredangle A=measuredangle C$.
Easy to see that $ABCD$ is not parallelogram.
answered Jan 8 at 12:54
Michael RozenbergMichael Rozenberg
100k1591192
$endgroup$
$begingroup$
$ABCD$ will always turn out to be concave?( I mean, I get a concave quadrilateral, and the nomenclature also differs, thats why)
$endgroup$
– Yellow
Jan 8 at 15:40
$begingroup$
And also, can you explain how $angle A$ = $angle C$?
$endgroup$
– Yellow
Jan 8 at 15:41
$begingroup$
@Anu Radha Yes, it concave. The equality of angles follows from congruent triangles.
$endgroup$
– Michael Rozenberg
Jan 8 at 16:29
$begingroup$
Is there any particular reason why the resulting quadrilateral is concave rather than being convex? Why can it NOT be convex?
$endgroup$
– Yellow
Jan 8 at 17:09
$begingroup$
Also, by $angle A$ and $angle C$, which two angles do you mean? I dont understand how equality follows from the congruency..(Sorry, really!)
$endgroup$
– Yellow
Jan 8 at 18:39
|
show 2 more comments
$begingroup$
Take $Delta ABK$ such that $measuredangle ABK>90^{circ}$ and let $Din AK$ such that $BD=BK$.
Now, let $Delta BDCcongDelta KBA$ such that $K$ and $C$ are placed in the same side respect to the line $DB$.
Thus, $ABCD$ is a needed quadrilateral because $AB=DC$ and $measuredangle A=measuredangle C$.
Easy to see that $ABCD$ is not parallelogram.
answered Jan 8 at 12:54
Michael RozenbergMichael Rozenberg
100k1591192
$endgroup$
Take $Delta ABK$ such that $measuredangle ABK>90^{circ}$ and let $Din AK$ such that $BD=BK$.
Now, let $Delta BDCcongDelta KBA$ such that $K$ and $C$ are placed in the same side respect to the line $DB$.
Thus, $ABCD$ is a needed quadrilateral because $AB=DC$ and $measuredangle A=measuredangle C$.
Easy to see that $ABCD$ is not parallelogram.
answered Jan 8 at 12:54
Michael RozenbergMichael Rozenberg
100k1591192
answered Jan 8 at 12:54
Michael RozenbergMichael Rozenberg
100k1591192
answered Jan 8 at 12:54
Michael RozenbergMichael Rozenberg
100k1591192
answered Jan 8 at 12:54
Michael RozenbergMichael Rozenberg
100k1591192
100k1591192
$begingroup$
$ABCD$ will always turn out to be concave?( I mean, I get a concave quadrilateral, and the nomenclature also differs, thats why)
$endgroup$
– Yellow
Jan 8 at 15:40
$begingroup$
And also, can you explain how $angle A$ = $angle C$?
$endgroup$
– Yellow
Jan 8 at 15:41
$begingroup$
@Anu Radha Yes, it concave. The equality of angles follows from congruent triangles.
$endgroup$
– Michael Rozenberg
Jan 8 at 16:29
$begingroup$
Is there any particular reason why the resulting quadrilateral is concave rather than being convex? Why can it NOT be convex?
$endgroup$
– Yellow
Jan 8 at 17:09
$begingroup$
Also, by $angle A$ and $angle C$, which two angles do you mean? I dont understand how equality follows from the congruency..(Sorry, really!)
$endgroup$
– Yellow
Jan 8 at 18:39
|
show 2 more comments
$begingroup$
$ABCD$ will always turn out to be concave?( I mean, I get a concave quadrilateral, and the nomenclature also differs, thats why)
$endgroup$
– Yellow
Jan 8 at 15:40
$begingroup$
And also, can you explain how $angle A$ = $angle C$?
$endgroup$
– Yellow
Jan 8 at 15:41
$begingroup$
@Anu Radha Yes, it concave. The equality of angles follows from congruent triangles.
$endgroup$
– Michael Rozenberg
Jan 8 at 16:29
$begingroup$
Is there any particular reason why the resulting quadrilateral is concave rather than being convex? Why can it NOT be convex?
$endgroup$
– Yellow
Jan 8 at 17:09
$begingroup$
Also, by $angle A$ and $angle C$, which two angles do you mean? I dont understand how equality follows from the congruency..(Sorry, really!)
$endgroup$
– Yellow
Jan 8 at 18:39
$begingroup$
$ABCD$ will always turn out to be concave?( I mean, I get a concave quadrilateral, and the nomenclature also differs, thats why)
$endgroup$
– Yellow
Jan 8 at 15:40
$begingroup$
$ABCD$ will always turn out to be concave?( I mean, I get a concave quadrilateral, and the nomenclature also differs, thats why)
$endgroup$
– Yellow
Jan 8 at 15:40
$begingroup$
And also, can you explain how $angle A$ = $angle C$?
$endgroup$
– Yellow
Jan 8 at 15:41
$begingroup$
And also, can you explain how $angle A$ = $angle C$?
$endgroup$
– Yellow
Jan 8 at 15:41
$begingroup$
@Anu Radha Yes, it concave. The equality of angles follows from congruent triangles.
$endgroup$
– Michael Rozenberg
Jan 8 at 16:29
$begingroup$
@Anu Radha Yes, it concave. The equality of angles follows from congruent triangles.
$endgroup$
– Michael Rozenberg
Jan 8 at 16:29
$begingroup$
Is there any particular reason why the resulting quadrilateral is concave rather than being convex? Why can it NOT be convex?
$endgroup$
– Yellow
Jan 8 at 17:09
$begingroup$
Is there any particular reason why the resulting quadrilateral is concave rather than being convex? Why can it NOT be convex?
$endgroup$
– Yellow
Jan 8 at 17:09
$begingroup$
Also, by $angle A$ and $angle C$, which two angles do you mean? I dont understand how equality follows from the congruency..(Sorry, really!)
$endgroup$
– Yellow
Jan 8 at 18:39
$begingroup$
Also, by $angle A$ and $angle C$, which two angles do you mean? I dont understand how equality follows from the congruency..(Sorry, really!)
$endgroup$
– Yellow
Jan 8 at 18:39
|
show 2 more comments
$begingroup$
You are not succeeding in finding such a quadrilateral because none exists.
If $ABCD$ is such a quadrilateral with $|AD| = |BC|$ and $beta = delta$ then the triangles $ABC$ and $CDA$ are congruent. It follows that $sphericalangle CAB = sphericalangle DCA$, hence $AB$ is parallel to $DC$ from this you can easily deduce that the quadrilateral is a parallelogram.
answered Jan 8 at 12:53
Sergey YurkevichSergey Yurkevich
162
$endgroup$
$begingroup$
They are not always congruent. See please my counterexample.
$endgroup$
– Michael Rozenberg
Jan 8 at 13:06
add a comment |
$begingroup$
You are not succeeding in finding such a quadrilateral because none exists.
If $ABCD$ is such a quadrilateral with $|AD| = |BC|$ and $beta = delta$ then the triangles $ABC$ and $CDA$ are congruent. It follows that $sphericalangle CAB = sphericalangle DCA$, hence $AB$ is parallel to $DC$ from this you can easily deduce that the quadrilateral is a parallelogram.
answered Jan 8 at 12:53
Sergey YurkevichSergey Yurkevich
162
$endgroup$
$begingroup$
They are not always congruent. See please my counterexample.
$endgroup$
– Michael Rozenberg
Jan 8 at 13:06
add a comment |
$begingroup$
You are not succeeding in finding such a quadrilateral because none exists.
If $ABCD$ is such a quadrilateral with $|AD| = |BC|$ and $beta = delta$ then the triangles $ABC$ and $CDA$ are congruent. It follows that $sphericalangle CAB = sphericalangle DCA$, hence $AB$ is parallel to $DC$ from this you can easily deduce that the quadrilateral is a parallelogram.
answered Jan 8 at 12:53
Sergey YurkevichSergey Yurkevich
162
$endgroup$
You are not succeeding in finding such a quadrilateral because none exists.
If $ABCD$ is such a quadrilateral with $|AD| = |BC|$ and $beta = delta$ then the triangles $ABC$ and $CDA$ are congruent. It follows that $sphericalangle CAB = sphericalangle DCA$, hence $AB$ is parallel to $DC$ from this you can easily deduce that the quadrilateral is a parallelogram.
answered Jan 8 at 12:53
Sergey YurkevichSergey Yurkevich
162
answered Jan 8 at 12:53
Sergey YurkevichSergey Yurkevich
162
answered Jan 8 at 12:53
Sergey YurkevichSergey Yurkevich
162
answered Jan 8 at 12:53
Sergey YurkevichSergey Yurkevich
162
162
$begingroup$
They are not always congruent. See please my counterexample.
$endgroup$
– Michael Rozenberg
Jan 8 at 13:06
add a comment |
$begingroup$
They are not always congruent. See please my counterexample.
$endgroup$
– Michael Rozenberg
Jan 8 at 13:06
$begingroup$
They are not always congruent. See please my counterexample.
$endgroup$
– Michael Rozenberg
Jan 8 at 13:06
$begingroup$
They are not always congruent. See please my counterexample.
$endgroup$
– Michael Rozenberg
Jan 8 at 13:06
add a comment |
$begingroup$
Related/Duplicate: "If a quadrilateral has a pair of equal opposite sides, and a pair of equal opposite angles, then is it necessarily a parallelogram?" and "Construct or prove existence of a certain quadrilateral". Perhaps others.
$endgroup$
– Blue
Jan 8 at 13:28
$begingroup$
Is it fine now?
$endgroup$
– Yellow
Jan 8 at 15:24