Relationship between two traingles that have same adjacent side
If we have two triangles $akz$ and $bkc$ in which they have same side $k$.
If $t_1$ and $t_2$ are two angles $in thinspace]0thinspace , thinspace 90[$ in which the above-mentioned $k$ is the adjacent as shown in the following image:
two triangles illustration
Can we prove that:
if $b > a > k$ $Rightarrow$ $c > z$ is always true?
geometry proof-writing triangle
asked Nov 25 '18 at 11:15
Mike
54
add a comment |
If we have two triangles $akz$ and $bkc$ in which they have same side $k$.
If $t_1$ and $t_2$ are two angles $in thinspace]0thinspace , thinspace 90[$ in which the above-mentioned $k$ is the adjacent as shown in the following image:
two triangles illustration
Can we prove that:
if $b > a > k$ $Rightarrow$ $c > z$ is always true?
geometry proof-writing triangle
asked Nov 25 '18 at 11:15
Mike
54
add a comment |
If we have two triangles $akz$ and $bkc$ in which they have same side $k$.
If $t_1$ and $t_2$ are two angles $in thinspace]0thinspace , thinspace 90[$ in which the above-mentioned $k$ is the adjacent as shown in the following image:
two triangles illustration
Can we prove that:
if $b > a > k$ $Rightarrow$ $c > z$ is always true?
geometry proof-writing triangle
asked Nov 25 '18 at 11:15
Mike
54
If we have two triangles $akz$ and $bkc$ in which they have same side $k$.
If $t_1$ and $t_2$ are two angles $in thinspace]0thinspace , thinspace 90[$ in which the above-mentioned $k$ is the adjacent as shown in the following image:
two triangles illustration
Can we prove that:
if $b > a > k$ $Rightarrow$ $c > z$ is always true?
geometry proof-writing triangle
geometry proof-writing triangle
asked Nov 25 '18 at 11:15
Mike
54
asked Nov 25 '18 at 11:15
Mike
54
edited Nov 25 '18 at 11:30
asked Nov 25 '18 at 11:15
Mike
54
asked Nov 25 '18 at 11:15
Mike
54
asked Nov 25 '18 at 11:15
Mike
54
54
add a comment |
add a comment |
1 Answer
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No. $X = (0, 0)$, $Y = (4, 0)$ and $Z = (4, 2)$, and
$W = (5, 1)$. Here $k = [X, Y]$, $a = [X, Z]$, $ z = [Y, Z]$ and
$ b = [X, W]$ and $ c = [Y, W]$. Note, $|k| = 4$, $|a| = sqrt{20}$ $|z| = 2$,
$|b| = sqrt{26}$, and $|c| = sqrt{2}$.
Geometrywise, $b$ is outside the circle of radius $|a|$ and centre $X$ (not too far) and much nearer the line $k$.
answered Nov 25 '18 at 12:04
R.C.Cowsik
31514
Thanks for your answer, before accepting it I just wanna make sure: Is that also applicable in $3D$?
– Mike
Nov 25 '18 at 12:37
Here is a link of your example, you can take a screenshot and add it to your answer
– Mike
Nov 25 '18 at 12:39
add a comment |
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1 Answer
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1 Answer
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No. $X = (0, 0)$, $Y = (4, 0)$ and $Z = (4, 2)$, and
$W = (5, 1)$. Here $k = [X, Y]$, $a = [X, Z]$, $ z = [Y, Z]$ and
$ b = [X, W]$ and $ c = [Y, W]$. Note, $|k| = 4$, $|a| = sqrt{20}$ $|z| = 2$,
$|b| = sqrt{26}$, and $|c| = sqrt{2}$.
Geometrywise, $b$ is outside the circle of radius $|a|$ and centre $X$ (not too far) and much nearer the line $k$.
answered Nov 25 '18 at 12:04
R.C.Cowsik
31514
Thanks for your answer, before accepting it I just wanna make sure: Is that also applicable in $3D$?
– Mike
Nov 25 '18 at 12:37
Here is a link of your example, you can take a screenshot and add it to your answer
– Mike
Nov 25 '18 at 12:39
add a comment |
No. $X = (0, 0)$, $Y = (4, 0)$ and $Z = (4, 2)$, and
$W = (5, 1)$. Here $k = [X, Y]$, $a = [X, Z]$, $ z = [Y, Z]$ and
$ b = [X, W]$ and $ c = [Y, W]$. Note, $|k| = 4$, $|a| = sqrt{20}$ $|z| = 2$,
$|b| = sqrt{26}$, and $|c| = sqrt{2}$.
Geometrywise, $b$ is outside the circle of radius $|a|$ and centre $X$ (not too far) and much nearer the line $k$.
answered Nov 25 '18 at 12:04
R.C.Cowsik
31514
Thanks for your answer, before accepting it I just wanna make sure: Is that also applicable in $3D$?
– Mike
Nov 25 '18 at 12:37
Here is a link of your example, you can take a screenshot and add it to your answer
– Mike
Nov 25 '18 at 12:39
add a comment |
No. $X = (0, 0)$, $Y = (4, 0)$ and $Z = (4, 2)$, and
$W = (5, 1)$. Here $k = [X, Y]$, $a = [X, Z]$, $ z = [Y, Z]$ and
$ b = [X, W]$ and $ c = [Y, W]$. Note, $|k| = 4$, $|a| = sqrt{20}$ $|z| = 2$,
$|b| = sqrt{26}$, and $|c| = sqrt{2}$.
Geometrywise, $b$ is outside the circle of radius $|a|$ and centre $X$ (not too far) and much nearer the line $k$.
answered Nov 25 '18 at 12:04
R.C.Cowsik
31514
No. $X = (0, 0)$, $Y = (4, 0)$ and $Z = (4, 2)$, and
$W = (5, 1)$. Here $k = [X, Y]$, $a = [X, Z]$, $ z = [Y, Z]$ and
$ b = [X, W]$ and $ c = [Y, W]$. Note, $|k| = 4$, $|a| = sqrt{20}$ $|z| = 2$,
$|b| = sqrt{26}$, and $|c| = sqrt{2}$.
Geometrywise, $b$ is outside the circle of radius $|a|$ and centre $X$ (not too far) and much nearer the line $k$.
answered Nov 25 '18 at 12:04
R.C.Cowsik
31514
answered Nov 25 '18 at 12:04
R.C.Cowsik
31514
answered Nov 25 '18 at 12:04
R.C.Cowsik
31514
answered Nov 25 '18 at 12:04
R.C.Cowsik
31514
31514
Thanks for your answer, before accepting it I just wanna make sure: Is that also applicable in $3D$?
– Mike
Nov 25 '18 at 12:37
Here is a link of your example, you can take a screenshot and add it to your answer
– Mike
Nov 25 '18 at 12:39
add a comment |
Thanks for your answer, before accepting it I just wanna make sure: Is that also applicable in $3D$?
– Mike
Nov 25 '18 at 12:37
Here is a link of your example, you can take a screenshot and add it to your answer
– Mike
Nov 25 '18 at 12:39
Thanks for your answer, before accepting it I just wanna make sure: Is that also applicable in $3D$?
– Mike
Nov 25 '18 at 12:37
Thanks for your answer, before accepting it I just wanna make sure: Is that also applicable in $3D$?
– Mike
Nov 25 '18 at 12:37
Here is a link of your example, you can take a screenshot and add it to your answer
– Mike
Nov 25 '18 at 12:39
Here is a link of your example, you can take a screenshot and add it to your answer
– Mike
Nov 25 '18 at 12:39
add a comment |
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