Impossible to have a triangle where trisectors of an angle trisect the opposite side
$begingroup$
Prove that there cannot exist a triangle in which the trisectors of an angle also trisect the opposite side by using proportions.
Attempt
I have started the proof by way of contradiction.
Suppose we have a triangle $Delta BAC$ with trisectors $AD$ and $A$E so that $D$ and $E$ are points lying in side $BC$. I also supposed that $D$ and $E$ trisect $BC$ such that $BD=DE=EC$. I then looked at triangle $BAE$ and observed that $AD$ bisects $angle BAE$ and that point $D$ is the midpoint of $BE$ therefore $BD$ is both a median and an angle bisector.
From this I get stuck and do not know how to apply proportions. I know we proceed similarly when looking at $Delta DAC$.
geometry triangle
edited Dec 19 '18 at 21:36
jayant98
657318
asked Dec 19 '18 at 19:27
dhuert4 dhuert4
12
$endgroup$
add a comment |
$begingroup$
Prove that there cannot exist a triangle in which the trisectors of an angle also trisect the opposite side by using proportions.
Attempt
I have started the proof by way of contradiction.
Suppose we have a triangle $Delta BAC$ with trisectors $AD$ and $A$E so that $D$ and $E$ are points lying in side $BC$. I also supposed that $D$ and $E$ trisect $BC$ such that $BD=DE=EC$. I then looked at triangle $BAE$ and observed that $AD$ bisects $angle BAE$ and that point $D$ is the midpoint of $BE$ therefore $BD$ is both a median and an angle bisector.
From this I get stuck and do not know how to apply proportions. I know we proceed similarly when looking at $Delta DAC$.
geometry triangle
edited Dec 19 '18 at 21:36
jayant98
657318
asked Dec 19 '18 at 19:27
dhuert4 dhuert4
12
$endgroup$
add a comment |
$begingroup$
Prove that there cannot exist a triangle in which the trisectors of an angle also trisect the opposite side by using proportions.
Attempt
I have started the proof by way of contradiction.
Suppose we have a triangle $Delta BAC$ with trisectors $AD$ and $A$E so that $D$ and $E$ are points lying in side $BC$. I also supposed that $D$ and $E$ trisect $BC$ such that $BD=DE=EC$. I then looked at triangle $BAE$ and observed that $AD$ bisects $angle BAE$ and that point $D$ is the midpoint of $BE$ therefore $BD$ is both a median and an angle bisector.
From this I get stuck and do not know how to apply proportions. I know we proceed similarly when looking at $Delta DAC$.
geometry triangle
edited Dec 19 '18 at 21:36
jayant98
657318
asked Dec 19 '18 at 19:27
dhuert4 dhuert4
12
$endgroup$
Prove that there cannot exist a triangle in which the trisectors of an angle also trisect the opposite side by using proportions.
Attempt
I have started the proof by way of contradiction.
Suppose we have a triangle $Delta BAC$ with trisectors $AD$ and $A$E so that $D$ and $E$ are points lying in side $BC$. I also supposed that $D$ and $E$ trisect $BC$ such that $BD=DE=EC$. I then looked at triangle $BAE$ and observed that $AD$ bisects $angle BAE$ and that point $D$ is the midpoint of $BE$ therefore $BD$ is both a median and an angle bisector.
From this I get stuck and do not know how to apply proportions. I know we proceed similarly when looking at $Delta DAC$.
geometry triangle
geometry triangle
edited Dec 19 '18 at 21:36
jayant98
657318
asked Dec 19 '18 at 19:27
dhuert4 dhuert4
12
edited Dec 19 '18 at 21:36
jayant98
657318
asked Dec 19 '18 at 19:27
dhuert4 dhuert4
12
edited Dec 19 '18 at 21:36
jayant98
657318
edited Dec 19 '18 at 21:36
jayant98
657318
edited Dec 19 '18 at 21:36
jayant98
657318
657318
asked Dec 19 '18 at 19:27
dhuert4 dhuert4
12
asked Dec 19 '18 at 19:27
dhuert4 dhuert4
12
asked Dec 19 '18 at 19:27
dhuert4 dhuert4
12
12
add a comment |
add a comment |
2 Answers
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votes
$begingroup$
It is enough to show (through the sine theorem, for instance) that in the following configuration the red segments are longer than the blue ones:
answered Dec 19 '18 at 19:45
Jack D'AurizioJack D'Aurizio
290k33283664
$endgroup$
add a comment |
$begingroup$
You have a $triangle{ABC}$. Let us say that side $BC$ is trisected by points $D, E$ such as $BD=DE=EC$. Note that $A_{triangle{ACE}}=A_{triangle{AED}}=A_{triangle{ADB}}$ (equal bases, same height). Using area of a triangle formula this leads to $AC=AD$, $AE=AB$. But then $AE$ is a median and height of $triangle{ACD}$ and $AD$ is a median and height of $triangle{AED}$. We have two perpediculars from one point $A$ to side $BC$ that go to two distinct points. This is impossible.
answered Dec 19 '18 at 19:51
VasyaVasya
3,3871516
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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votes
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votes
$begingroup$
It is enough to show (through the sine theorem, for instance) that in the following configuration the red segments are longer than the blue ones:
answered Dec 19 '18 at 19:45
Jack D'AurizioJack D'Aurizio
290k33283664
$endgroup$
add a comment |
$begingroup$
It is enough to show (through the sine theorem, for instance) that in the following configuration the red segments are longer than the blue ones:
answered Dec 19 '18 at 19:45
Jack D'AurizioJack D'Aurizio
290k33283664
$endgroup$
add a comment |
$begingroup$
It is enough to show (through the sine theorem, for instance) that in the following configuration the red segments are longer than the blue ones:
answered Dec 19 '18 at 19:45
Jack D'AurizioJack D'Aurizio
290k33283664
$endgroup$
It is enough to show (through the sine theorem, for instance) that in the following configuration the red segments are longer than the blue ones:
answered Dec 19 '18 at 19:45
Jack D'AurizioJack D'Aurizio
290k33283664
answered Dec 19 '18 at 19:45
Jack D'AurizioJack D'Aurizio
290k33283664
answered Dec 19 '18 at 19:45
Jack D'AurizioJack D'Aurizio
290k33283664
answered Dec 19 '18 at 19:45
Jack D'AurizioJack D'Aurizio
290k33283664
290k33283664
add a comment |
add a comment |
$begingroup$
You have a $triangle{ABC}$. Let us say that side $BC$ is trisected by points $D, E$ such as $BD=DE=EC$. Note that $A_{triangle{ACE}}=A_{triangle{AED}}=A_{triangle{ADB}}$ (equal bases, same height). Using area of a triangle formula this leads to $AC=AD$, $AE=AB$. But then $AE$ is a median and height of $triangle{ACD}$ and $AD$ is a median and height of $triangle{AED}$. We have two perpediculars from one point $A$ to side $BC$ that go to two distinct points. This is impossible.
answered Dec 19 '18 at 19:51
VasyaVasya
3,3871516
$endgroup$
add a comment |
$begingroup$
You have a $triangle{ABC}$. Let us say that side $BC$ is trisected by points $D, E$ such as $BD=DE=EC$. Note that $A_{triangle{ACE}}=A_{triangle{AED}}=A_{triangle{ADB}}$ (equal bases, same height). Using area of a triangle formula this leads to $AC=AD$, $AE=AB$. But then $AE$ is a median and height of $triangle{ACD}$ and $AD$ is a median and height of $triangle{AED}$. We have two perpediculars from one point $A$ to side $BC$ that go to two distinct points. This is impossible.
answered Dec 19 '18 at 19:51
VasyaVasya
3,3871516
$endgroup$
add a comment |
$begingroup$
You have a $triangle{ABC}$. Let us say that side $BC$ is trisected by points $D, E$ such as $BD=DE=EC$. Note that $A_{triangle{ACE}}=A_{triangle{AED}}=A_{triangle{ADB}}$ (equal bases, same height). Using area of a triangle formula this leads to $AC=AD$, $AE=AB$. But then $AE$ is a median and height of $triangle{ACD}$ and $AD$ is a median and height of $triangle{AED}$. We have two perpediculars from one point $A$ to side $BC$ that go to two distinct points. This is impossible.
answered Dec 19 '18 at 19:51
VasyaVasya
3,3871516
$endgroup$
You have a $triangle{ABC}$. Let us say that side $BC$ is trisected by points $D, E$ such as $BD=DE=EC$. Note that $A_{triangle{ACE}}=A_{triangle{AED}}=A_{triangle{ADB}}$ (equal bases, same height). Using area of a triangle formula this leads to $AC=AD$, $AE=AB$. But then $AE$ is a median and height of $triangle{ACD}$ and $AD$ is a median and height of $triangle{AED}$. We have two perpediculars from one point $A$ to side $BC$ that go to two distinct points. This is impossible.
answered Dec 19 '18 at 19:51
VasyaVasya
3,3871516
answered Dec 19 '18 at 19:51
VasyaVasya
3,3871516
answered Dec 19 '18 at 19:51
VasyaVasya
3,3871516
answered Dec 19 '18 at 19:51
VasyaVasya
3,3871516
3,3871516
add a comment |
add a comment |
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