Minimum steps to prove four points form a square. Announcing the arrival of Valued Associate...
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Minimum steps to prove four points form a square.
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Given two diagonally opposite points on a square, how to calculate the other two pointsShow centers of squares formed by a parallelogram form a square.How to prove the quadrilateral formed by bisectors of a parallelogram is not always square?Given two adjacent sides of a rectangle are equivalent, prove that the quadrilateral is a square.Proving $angle QAP=45^circ$ if $ABCD$ is a square with points $P$ in $BC$, $Q$ in $CD$ satisfying $overline{BP}+overline{DQ}=overline{PQ}$Maximum Perimeter of triangle in a SquareShow that a line l intersects all four sides of a quadrilateral given that none of the four points ABCD lies on l.How to compute the intercept of a square with a line given by two points?prove ABCD is also a squareGeometry/Trigonometry Question on two congruent isosceles triangles inside and outside a square.
$begingroup$
I have points $A, B, C, D$ and I need to show that $ABCD$ is a square. I tried this by showing all four sides have equal length and that two angles sharing a side are $90^{circ}$ degrees. Is there a more concise way to prove the points form a square?
geometry euclidean-geometry
edited Mar 26 at 11:08
Michael Rozenberg
111k1897201
asked Mar 26 at 10:59
Tom FinetTom Finet
9318
$endgroup$
add a comment |
$begingroup$
I have points $A, B, C, D$ and I need to show that $ABCD$ is a square. I tried this by showing all four sides have equal length and that two angles sharing a side are $90^{circ}$ degrees. Is there a more concise way to prove the points form a square?
geometry euclidean-geometry
edited Mar 26 at 11:08
Michael Rozenberg
111k1897201
asked Mar 26 at 10:59
Tom FinetTom Finet
9318
$endgroup$
$begingroup$
Diagonals of a square are equal. All sides are equal. Diagonals bisect each other at $90^{circ}$.
$endgroup$
– Paras Khosla
Mar 26 at 11:02
$begingroup$
One point finer than yours is that if the shared side is BC you only have to show AB BC and CD are equal (unless the shape can be self intersecting).
$endgroup$
– Paul Childs
Mar 26 at 11:39
add a comment |
$begingroup$
I have points $A, B, C, D$ and I need to show that $ABCD$ is a square. I tried this by showing all four sides have equal length and that two angles sharing a side are $90^{circ}$ degrees. Is there a more concise way to prove the points form a square?
geometry euclidean-geometry
edited Mar 26 at 11:08
Michael Rozenberg
111k1897201
asked Mar 26 at 10:59
Tom FinetTom Finet
9318
$endgroup$
I have points $A, B, C, D$ and I need to show that $ABCD$ is a square. I tried this by showing all four sides have equal length and that two angles sharing a side are $90^{circ}$ degrees. Is there a more concise way to prove the points form a square?
geometry euclidean-geometry
geometry euclidean-geometry
edited Mar 26 at 11:08
Michael Rozenberg
111k1897201
asked Mar 26 at 10:59
Tom FinetTom Finet
9318
edited Mar 26 at 11:08
Michael Rozenberg
111k1897201
asked Mar 26 at 10:59
Tom FinetTom Finet
9318
edited Mar 26 at 11:08
Michael Rozenberg
111k1897201
edited Mar 26 at 11:08
Michael Rozenberg
111k1897201
edited Mar 26 at 11:08
Michael Rozenberg
111k1897201
111k1897201
asked Mar 26 at 10:59
Tom FinetTom Finet
9318
asked Mar 26 at 10:59
Tom FinetTom Finet
9318
asked Mar 26 at 10:59
Tom FinetTom Finet
9318
9318
$begingroup$
Diagonals of a square are equal. All sides are equal. Diagonals bisect each other at $90^{circ}$.
$endgroup$
– Paras Khosla
Mar 26 at 11:02
$begingroup$
One point finer than yours is that if the shared side is BC you only have to show AB BC and CD are equal (unless the shape can be self intersecting).
$endgroup$
– Paul Childs
Mar 26 at 11:39
add a comment |
$begingroup$
Diagonals of a square are equal. All sides are equal. Diagonals bisect each other at $90^{circ}$.
$endgroup$
– Paras Khosla
Mar 26 at 11:02
$begingroup$
One point finer than yours is that if the shared side is BC you only have to show AB BC and CD are equal (unless the shape can be self intersecting).
$endgroup$
– Paul Childs
Mar 26 at 11:39
$begingroup$
Diagonals of a square are equal. All sides are equal. Diagonals bisect each other at $90^{circ}$.
$endgroup$
– Paras Khosla
Mar 26 at 11:02
$begingroup$
Diagonals of a square are equal. All sides are equal. Diagonals bisect each other at $90^{circ}$.
$endgroup$
– Paras Khosla
Mar 26 at 11:02
$begingroup$
One point finer than yours is that if the shared side is BC you only have to show AB BC and CD are equal (unless the shape can be self intersecting).
$endgroup$
– Paul Childs
Mar 26 at 11:39
$begingroup$
One point finer than yours is that if the shared side is BC you only have to show AB BC and CD are equal (unless the shape can be self intersecting).
$endgroup$
– Paul Childs
Mar 26 at 11:39
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Prove that $ACperp BD$, $AC=BD$ and $AC$ and $BD$ have the same midpoint.
answered Mar 26 at 11:03
Michael RozenbergMichael Rozenberg
111k1897201
$endgroup$
1
$begingroup$
Should be AC = BD. You also need to have the converse re bisection, namely that BD bisects AC (they aren't equivalent).
$endgroup$
– Paul Childs
Mar 26 at 11:35
$begingroup$
Thank you, @Paul ! I fixed.
$endgroup$
– Michael Rozenberg
Mar 26 at 11:56
$begingroup$
Awesome. But don't forget the second part, else a kite may satisfy your conditions. AC must bisect BD and BD must bisect AC.
$endgroup$
– Paul Childs
Mar 26 at 12:05
$begingroup$
It was my delirium. Thank you! I fixed again.
$endgroup$
– Michael Rozenberg
Mar 26 at 12:10
add a comment |
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1 Answer
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1 Answer
1
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oldest
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active
oldest
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active
oldest
votes
$begingroup$
Prove that $ACperp BD$, $AC=BD$ and $AC$ and $BD$ have the same midpoint.
answered Mar 26 at 11:03
Michael RozenbergMichael Rozenberg
111k1897201
$endgroup$
1
$begingroup$
Should be AC = BD. You also need to have the converse re bisection, namely that BD bisects AC (they aren't equivalent).
$endgroup$
– Paul Childs
Mar 26 at 11:35
$begingroup$
Thank you, @Paul ! I fixed.
$endgroup$
– Michael Rozenberg
Mar 26 at 11:56
$begingroup$
Awesome. But don't forget the second part, else a kite may satisfy your conditions. AC must bisect BD and BD must bisect AC.
$endgroup$
– Paul Childs
Mar 26 at 12:05
$begingroup$
It was my delirium. Thank you! I fixed again.
$endgroup$
– Michael Rozenberg
Mar 26 at 12:10
add a comment |
$begingroup$
Prove that $ACperp BD$, $AC=BD$ and $AC$ and $BD$ have the same midpoint.
answered Mar 26 at 11:03
Michael RozenbergMichael Rozenberg
111k1897201
$endgroup$
1
$begingroup$
Should be AC = BD. You also need to have the converse re bisection, namely that BD bisects AC (they aren't equivalent).
$endgroup$
– Paul Childs
Mar 26 at 11:35
$begingroup$
Thank you, @Paul ! I fixed.
$endgroup$
– Michael Rozenberg
Mar 26 at 11:56
$begingroup$
Awesome. But don't forget the second part, else a kite may satisfy your conditions. AC must bisect BD and BD must bisect AC.
$endgroup$
– Paul Childs
Mar 26 at 12:05
$begingroup$
It was my delirium. Thank you! I fixed again.
$endgroup$
– Michael Rozenberg
Mar 26 at 12:10
add a comment |
$begingroup$
Prove that $ACperp BD$, $AC=BD$ and $AC$ and $BD$ have the same midpoint.
answered Mar 26 at 11:03
Michael RozenbergMichael Rozenberg
111k1897201
$endgroup$
Prove that $ACperp BD$, $AC=BD$ and $AC$ and $BD$ have the same midpoint.
answered Mar 26 at 11:03
Michael RozenbergMichael Rozenberg
111k1897201
edited Mar 26 at 12:09
answered Mar 26 at 11:03
Michael RozenbergMichael Rozenberg
111k1897201
answered Mar 26 at 11:03
Michael RozenbergMichael Rozenberg
111k1897201
answered Mar 26 at 11:03
Michael RozenbergMichael Rozenberg
111k1897201
111k1897201
1
$begingroup$
Should be AC = BD. You also need to have the converse re bisection, namely that BD bisects AC (they aren't equivalent).
$endgroup$
– Paul Childs
Mar 26 at 11:35
$begingroup$
Thank you, @Paul ! I fixed.
$endgroup$
– Michael Rozenberg
Mar 26 at 11:56
$begingroup$
Awesome. But don't forget the second part, else a kite may satisfy your conditions. AC must bisect BD and BD must bisect AC.
$endgroup$
– Paul Childs
Mar 26 at 12:05
$begingroup$
It was my delirium. Thank you! I fixed again.
$endgroup$
– Michael Rozenberg
Mar 26 at 12:10
add a comment |
1
$begingroup$
Should be AC = BD. You also need to have the converse re bisection, namely that BD bisects AC (they aren't equivalent).
$endgroup$
– Paul Childs
Mar 26 at 11:35
$begingroup$
Thank you, @Paul ! I fixed.
$endgroup$
– Michael Rozenberg
Mar 26 at 11:56
$begingroup$
Awesome. But don't forget the second part, else a kite may satisfy your conditions. AC must bisect BD and BD must bisect AC.
$endgroup$
– Paul Childs
Mar 26 at 12:05
$begingroup$
It was my delirium. Thank you! I fixed again.
$endgroup$
– Michael Rozenberg
Mar 26 at 12:10
1
1
$begingroup$
Should be AC = BD. You also need to have the converse re bisection, namely that BD bisects AC (they aren't equivalent).
$endgroup$
– Paul Childs
Mar 26 at 11:35
$begingroup$
Should be AC = BD. You also need to have the converse re bisection, namely that BD bisects AC (they aren't equivalent).
$endgroup$
– Paul Childs
Mar 26 at 11:35
$begingroup$
Thank you, @Paul ! I fixed.
$endgroup$
– Michael Rozenberg
Mar 26 at 11:56
$begingroup$
Thank you, @Paul ! I fixed.
$endgroup$
– Michael Rozenberg
Mar 26 at 11:56
$begingroup$
Awesome. But don't forget the second part, else a kite may satisfy your conditions. AC must bisect BD and BD must bisect AC.
$endgroup$
– Paul Childs
Mar 26 at 12:05
$begingroup$
Awesome. But don't forget the second part, else a kite may satisfy your conditions. AC must bisect BD and BD must bisect AC.
$endgroup$
– Paul Childs
Mar 26 at 12:05
$begingroup$
It was my delirium. Thank you! I fixed again.
$endgroup$
– Michael Rozenberg
Mar 26 at 12:10
$begingroup$
It was my delirium. Thank you! I fixed again.
$endgroup$
– Michael Rozenberg
Mar 26 at 12:10
add a comment |
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$begingroup$
Diagonals of a square are equal. All sides are equal. Diagonals bisect each other at $90^{circ}$.
$endgroup$
– Paras Khosla
Mar 26 at 11:02
$begingroup$
One point finer than yours is that if the shared side is BC you only have to show AB BC and CD are equal (unless the shape can be self intersecting).
$endgroup$
– Paul Childs
Mar 26 at 11:39