Does the equality of the sum of opposite sides in a quadrilateral necessarily imply the existence of an...
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I had come across a question in which we had to show that a given quadrilateral, if subjected under a condition, had an incircle. So, will it be sufficient to show that $a+b=c+d$, if $a,b,c,d$ are the sides of the quadrilateral? If yes, then would someone please tell me how the equality of the sum of opposite sides in a quadrilateral necessarily implies the existence of an inscribed circle?
geometry circle quadrilateral
edited Dec 28 '18 at 20:37
Michael Rozenberg
104k1892197
asked Dec 28 '18 at 18:53
Shashwat1337Shashwat1337
789
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add a comment |
$begingroup$
I had come across a question in which we had to show that a given quadrilateral, if subjected under a condition, had an incircle. So, will it be sufficient to show that $a+b=c+d$, if $a,b,c,d$ are the sides of the quadrilateral? If yes, then would someone please tell me how the equality of the sum of opposite sides in a quadrilateral necessarily implies the existence of an inscribed circle?
geometry circle quadrilateral
edited Dec 28 '18 at 20:37
Michael Rozenberg
104k1892197
asked Dec 28 '18 at 18:53
Shashwat1337Shashwat1337
789
$endgroup$
2
$begingroup$
Yes, if the quadrilateral is convex this condition is sufficient. For a proof see en.wikipedia.org/wiki/Tangential_quadrilateral
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– user376343
Dec 28 '18 at 18:57
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@user376343 Thanks
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– Shashwat1337
Dec 28 '18 at 19:01
add a comment |
$begingroup$
I had come across a question in which we had to show that a given quadrilateral, if subjected under a condition, had an incircle. So, will it be sufficient to show that $a+b=c+d$, if $a,b,c,d$ are the sides of the quadrilateral? If yes, then would someone please tell me how the equality of the sum of opposite sides in a quadrilateral necessarily implies the existence of an inscribed circle?
geometry circle quadrilateral
edited Dec 28 '18 at 20:37
Michael Rozenberg
104k1892197
asked Dec 28 '18 at 18:53
Shashwat1337Shashwat1337
789
$endgroup$
I had come across a question in which we had to show that a given quadrilateral, if subjected under a condition, had an incircle. So, will it be sufficient to show that $a+b=c+d$, if $a,b,c,d$ are the sides of the quadrilateral? If yes, then would someone please tell me how the equality of the sum of opposite sides in a quadrilateral necessarily implies the existence of an inscribed circle?
geometry circle quadrilateral
geometry circle quadrilateral
edited Dec 28 '18 at 20:37
Michael Rozenberg
104k1892197
asked Dec 28 '18 at 18:53
Shashwat1337Shashwat1337
789
edited Dec 28 '18 at 20:37
Michael Rozenberg
104k1892197
asked Dec 28 '18 at 18:53
Shashwat1337Shashwat1337
789
edited Dec 28 '18 at 20:37
Michael Rozenberg
104k1892197
edited Dec 28 '18 at 20:37
Michael Rozenberg
104k1892197
edited Dec 28 '18 at 20:37
Michael Rozenberg
104k1892197
104k1892197
asked Dec 28 '18 at 18:53
Shashwat1337Shashwat1337
789
asked Dec 28 '18 at 18:53
Shashwat1337Shashwat1337
789
asked Dec 28 '18 at 18:53
Shashwat1337Shashwat1337
789
789
2
$begingroup$
Yes, if the quadrilateral is convex this condition is sufficient. For a proof see en.wikipedia.org/wiki/Tangential_quadrilateral
$endgroup$
– user376343
Dec 28 '18 at 18:57
$begingroup$
@user376343 Thanks
$endgroup$
– Shashwat1337
Dec 28 '18 at 19:01
add a comment |
2
$begingroup$
Yes, if the quadrilateral is convex this condition is sufficient. For a proof see en.wikipedia.org/wiki/Tangential_quadrilateral
$endgroup$
– user376343
Dec 28 '18 at 18:57
$begingroup$
@user376343 Thanks
$endgroup$
– Shashwat1337
Dec 28 '18 at 19:01
2
2
$begingroup$
Yes, if the quadrilateral is convex this condition is sufficient. For a proof see en.wikipedia.org/wiki/Tangential_quadrilateral
$endgroup$
– user376343
Dec 28 '18 at 18:57
$begingroup$
Yes, if the quadrilateral is convex this condition is sufficient. For a proof see en.wikipedia.org/wiki/Tangential_quadrilateral
$endgroup$
– user376343
Dec 28 '18 at 18:57
$begingroup$
@user376343 Thanks
$endgroup$
– Shashwat1337
Dec 28 '18 at 19:01
$begingroup$
@user376343 Thanks
$endgroup$
– Shashwat1337
Dec 28 '18 at 19:01
add a comment |
1 Answer
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Yes, it's true.
Let $ABCD$ be an equilateral with $AB+CD=BC+AD$.
Also, let $AE$ and $BE$ be bisectors of $angle BAD$ and $angle ABC$ respectively.
Thus, the distances from $E$ to $AB$, $AD$ and $BC$ they are equal, which says that the circle with a center $E$ and with a radius is this distance, is touched to sides $AD$, $AB$ and $BC$.
Now, let $CD_1$ is a tangent to the circle, where $D_1$ is placed on the line $AD$.
Thus, $$AB+CD_1=BC+AD_1$$ and
$$AB+CD=BC+AD$$ from the given.
Hence, $$CD_1-CD=AD_1-AD$$ or
$$CD_1-CD=pm DD_1,$$ which by the triangle inequality is possible, when $Dequiv D_1$ and we are done!
answered Dec 28 '18 at 20:31
Michael RozenbergMichael Rozenberg
104k1892197
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$begingroup$
Thank you very much. The proof was great!
$endgroup$
– Shashwat1337
Dec 30 '18 at 17:37
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@Shashwat1337 You are welcome!
$endgroup$
– Michael Rozenberg
Dec 30 '18 at 17:38
add a comment |
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1 Answer
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$begingroup$
Yes, it's true.
Let $ABCD$ be an equilateral with $AB+CD=BC+AD$.
Also, let $AE$ and $BE$ be bisectors of $angle BAD$ and $angle ABC$ respectively.
Thus, the distances from $E$ to $AB$, $AD$ and $BC$ they are equal, which says that the circle with a center $E$ and with a radius is this distance, is touched to sides $AD$, $AB$ and $BC$.
Now, let $CD_1$ is a tangent to the circle, where $D_1$ is placed on the line $AD$.
Thus, $$AB+CD_1=BC+AD_1$$ and
$$AB+CD=BC+AD$$ from the given.
Hence, $$CD_1-CD=AD_1-AD$$ or
$$CD_1-CD=pm DD_1,$$ which by the triangle inequality is possible, when $Dequiv D_1$ and we are done!
answered Dec 28 '18 at 20:31
Michael RozenbergMichael Rozenberg
104k1892197
$endgroup$
$begingroup$
Thank you very much. The proof was great!
$endgroup$
– Shashwat1337
Dec 30 '18 at 17:37
$begingroup$
@Shashwat1337 You are welcome!
$endgroup$
– Michael Rozenberg
Dec 30 '18 at 17:38
add a comment |
$begingroup$
Yes, it's true.
Let $ABCD$ be an equilateral with $AB+CD=BC+AD$.
Also, let $AE$ and $BE$ be bisectors of $angle BAD$ and $angle ABC$ respectively.
Thus, the distances from $E$ to $AB$, $AD$ and $BC$ they are equal, which says that the circle with a center $E$ and with a radius is this distance, is touched to sides $AD$, $AB$ and $BC$.
Now, let $CD_1$ is a tangent to the circle, where $D_1$ is placed on the line $AD$.
Thus, $$AB+CD_1=BC+AD_1$$ and
$$AB+CD=BC+AD$$ from the given.
Hence, $$CD_1-CD=AD_1-AD$$ or
$$CD_1-CD=pm DD_1,$$ which by the triangle inequality is possible, when $Dequiv D_1$ and we are done!
answered Dec 28 '18 at 20:31
Michael RozenbergMichael Rozenberg
104k1892197
$endgroup$
$begingroup$
Thank you very much. The proof was great!
$endgroup$
– Shashwat1337
Dec 30 '18 at 17:37
$begingroup$
@Shashwat1337 You are welcome!
$endgroup$
– Michael Rozenberg
Dec 30 '18 at 17:38
add a comment |
$begingroup$
Yes, it's true.
Let $ABCD$ be an equilateral with $AB+CD=BC+AD$.
Also, let $AE$ and $BE$ be bisectors of $angle BAD$ and $angle ABC$ respectively.
Thus, the distances from $E$ to $AB$, $AD$ and $BC$ they are equal, which says that the circle with a center $E$ and with a radius is this distance, is touched to sides $AD$, $AB$ and $BC$.
Now, let $CD_1$ is a tangent to the circle, where $D_1$ is placed on the line $AD$.
Thus, $$AB+CD_1=BC+AD_1$$ and
$$AB+CD=BC+AD$$ from the given.
Hence, $$CD_1-CD=AD_1-AD$$ or
$$CD_1-CD=pm DD_1,$$ which by the triangle inequality is possible, when $Dequiv D_1$ and we are done!
answered Dec 28 '18 at 20:31
Michael RozenbergMichael Rozenberg
104k1892197
$endgroup$
Yes, it's true.
Let $ABCD$ be an equilateral with $AB+CD=BC+AD$.
Also, let $AE$ and $BE$ be bisectors of $angle BAD$ and $angle ABC$ respectively.
Thus, the distances from $E$ to $AB$, $AD$ and $BC$ they are equal, which says that the circle with a center $E$ and with a radius is this distance, is touched to sides $AD$, $AB$ and $BC$.
Now, let $CD_1$ is a tangent to the circle, where $D_1$ is placed on the line $AD$.
Thus, $$AB+CD_1=BC+AD_1$$ and
$$AB+CD=BC+AD$$ from the given.
Hence, $$CD_1-CD=AD_1-AD$$ or
$$CD_1-CD=pm DD_1,$$ which by the triangle inequality is possible, when $Dequiv D_1$ and we are done!
answered Dec 28 '18 at 20:31
Michael RozenbergMichael Rozenberg
104k1892197
answered Dec 28 '18 at 20:31
Michael RozenbergMichael Rozenberg
104k1892197
answered Dec 28 '18 at 20:31
Michael RozenbergMichael Rozenberg
104k1892197
answered Dec 28 '18 at 20:31
Michael RozenbergMichael Rozenberg
104k1892197
104k1892197
$begingroup$
Thank you very much. The proof was great!
$endgroup$
– Shashwat1337
Dec 30 '18 at 17:37
$begingroup$
@Shashwat1337 You are welcome!
$endgroup$
– Michael Rozenberg
Dec 30 '18 at 17:38
add a comment |
$begingroup$
Thank you very much. The proof was great!
$endgroup$
– Shashwat1337
Dec 30 '18 at 17:37
$begingroup$
@Shashwat1337 You are welcome!
$endgroup$
– Michael Rozenberg
Dec 30 '18 at 17:38
$begingroup$
Thank you very much. The proof was great!
$endgroup$
– Shashwat1337
Dec 30 '18 at 17:37
$begingroup$
Thank you very much. The proof was great!
$endgroup$
– Shashwat1337
Dec 30 '18 at 17:37
$begingroup$
@Shashwat1337 You are welcome!
$endgroup$
– Michael Rozenberg
Dec 30 '18 at 17:38
$begingroup$
@Shashwat1337 You are welcome!
$endgroup$
– Michael Rozenberg
Dec 30 '18 at 17:38
add a comment |
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2
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Yes, if the quadrilateral is convex this condition is sufficient. For a proof see en.wikipedia.org/wiki/Tangential_quadrilateral
$endgroup$
– user376343
Dec 28 '18 at 18:57
$begingroup$
@user376343 Thanks
$endgroup$
– Shashwat1337
Dec 28 '18 at 19:01