Prove $L$, $M$, $N$ are collinear The Next CEO of Stack OverflowProving a triangle is isoceles given two points on the sidesBisector of angle formed at the orthocentre passes through the circumcentreProve that L,M,N are collinear.Are lines which pass respectively through vertices $A,B,C$ and incenter, circumcenter and orthocenter of $Delta ABC$ concurrent?Collinearity of points in a projective settingWhy are auxiliary lines valid in geometric proofs?Finding the 'three points'Circumcenter of $Delta ABC$ is the centroid of the antipedal triangle of symmedian pointA conjecture related to any triangleShow that three point $G,H,G_1$ are collinear.
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Prove $L$, $M$, $N$ are collinear
The Next CEO of Stack OverflowProving a triangle is isoceles given two points on the sidesBisector of angle formed at the orthocentre passes through the circumcentreProve that L,M,N are collinear.Are lines which pass respectively through vertices $A,B,C$ and incenter, circumcenter and orthocenter of $Delta ABC$ concurrent?Collinearity of points in a projective settingWhy are auxiliary lines valid in geometric proofs?Finding the 'three points'Circumcenter of $Delta ABC$ is the centroid of the antipedal triangle of symmedian pointA conjecture related to any triangleShow that three point $G,H,G_1$ are collinear.
$begingroup$
$G$ is the centroid of triangle $ABC$; $AG$ is produced to $X$ such that $GX=AC$, if we draw parallels through $X$ to $CA$, $AB$, $BC$ meeting $BC$, $CA$, $AB$ at $L$, $M$, $N$ respectively, prove that $L$, $M$, $N$ are collinear.
I have tried using Menelaus theorem and thus tried to multiply the corresponding ratios to obtain $-1$ but I can't do so. I don't seem to understand the relevance of $GX=AC$ and that's probably why I can't solve it. Any hint would be appreciated, thank you.
geometry contest-math
edited 2 days ago
Anirban Niloy
8411318
asked 2 days ago
StompStomp
161
New contributor
Stomp is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
$G$ is the centroid of triangle $ABC$; $AG$ is produced to $X$ such that $GX=AC$, if we draw parallels through $X$ to $CA$, $AB$, $BC$ meeting $BC$, $CA$, $AB$ at $L$, $M$, $N$ respectively, prove that $L$, $M$, $N$ are collinear.
I have tried using Menelaus theorem and thus tried to multiply the corresponding ratios to obtain $-1$ but I can't do so. I don't seem to understand the relevance of $GX=AC$ and that's probably why I can't solve it. Any hint would be appreciated, thank you.
geometry contest-math
edited 2 days ago
Anirban Niloy
8411318
asked 2 days ago
StompStomp
161
New contributor
Stomp is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
I don't understand what "$AG$ is produced to $X$" means. Did you draw a picture? Can you share it with us?
$endgroup$
– Matteo
2 days ago
3
$begingroup$
Are you sure that $GX = AC$? I've calculated that you need $GX=AG$ for this theorem to be true.
$endgroup$
– Adam Latosiński
2 days ago
$begingroup$
Could you provide a picture??
$endgroup$
– Dr. Mathva
2 days ago
2
$begingroup$
@Matteo "Produce" is a slightly old-fashioned term in geometry meaning to extend or lengthen. You can find the definition in Wiktionary.
$endgroup$
– FredH
2 days ago
$begingroup$
Thanks you guys ,if AG is equal to GX then the problem becomes much easier.I think it was misprint that wasted a week of my time. thanks anyways
$endgroup$
– Stomp
2 days ago
add a comment |
$begingroup$
$G$ is the centroid of triangle $ABC$; $AG$ is produced to $X$ such that $GX=AC$, if we draw parallels through $X$ to $CA$, $AB$, $BC$ meeting $BC$, $CA$, $AB$ at $L$, $M$, $N$ respectively, prove that $L$, $M$, $N$ are collinear.
I have tried using Menelaus theorem and thus tried to multiply the corresponding ratios to obtain $-1$ but I can't do so. I don't seem to understand the relevance of $GX=AC$ and that's probably why I can't solve it. Any hint would be appreciated, thank you.
geometry contest-math
edited 2 days ago
Anirban Niloy
8411318
asked 2 days ago
StompStomp
161
New contributor
Stomp is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$G$ is the centroid of triangle $ABC$; $AG$ is produced to $X$ such that $GX=AC$, if we draw parallels through $X$ to $CA$, $AB$, $BC$ meeting $BC$, $CA$, $AB$ at $L$, $M$, $N$ respectively, prove that $L$, $M$, $N$ are collinear.
I have tried using Menelaus theorem and thus tried to multiply the corresponding ratios to obtain $-1$ but I can't do so. I don't seem to understand the relevance of $GX=AC$ and that's probably why I can't solve it. Any hint would be appreciated, thank you.
geometry contest-math
geometry contest-math
edited 2 days ago
Anirban Niloy
8411318
asked 2 days ago
StompStomp
161
New contributor
Stomp is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 days ago
Anirban Niloy
8411318
asked 2 days ago
StompStomp
161
New contributor
Stomp is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 days ago
Anirban Niloy
8411318
edited 2 days ago
Anirban Niloy
8411318
edited 2 days ago
Anirban Niloy
8411318
8411318
asked 2 days ago
StompStomp
161
New contributor
Stomp is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
StompStomp
161
asked 2 days ago
StompStomp
161
161
New contributor
Stomp is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Stomp is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Stomp is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
I don't understand what "$AG$ is produced to $X$" means. Did you draw a picture? Can you share it with us?
$endgroup$
– Matteo
2 days ago
3
$begingroup$
Are you sure that $GX = AC$? I've calculated that you need $GX=AG$ for this theorem to be true.
$endgroup$
– Adam Latosiński
2 days ago
$begingroup$
Could you provide a picture??
$endgroup$
– Dr. Mathva
2 days ago
2
$begingroup$
@Matteo "Produce" is a slightly old-fashioned term in geometry meaning to extend or lengthen. You can find the definition in Wiktionary.
$endgroup$
– FredH
2 days ago
$begingroup$
Thanks you guys ,if AG is equal to GX then the problem becomes much easier.I think it was misprint that wasted a week of my time. thanks anyways
$endgroup$
– Stomp
2 days ago
add a comment |
1
$begingroup$
I don't understand what "$AG$ is produced to $X$" means. Did you draw a picture? Can you share it with us?
$endgroup$
– Matteo
2 days ago
3
$begingroup$
Are you sure that $GX = AC$? I've calculated that you need $GX=AG$ for this theorem to be true.
$endgroup$
– Adam Latosiński
2 days ago
$begingroup$
Could you provide a picture??
$endgroup$
– Dr. Mathva
2 days ago
2
$begingroup$
@Matteo "Produce" is a slightly old-fashioned term in geometry meaning to extend or lengthen. You can find the definition in Wiktionary.
$endgroup$
– FredH
2 days ago
$begingroup$
Thanks you guys ,if AG is equal to GX then the problem becomes much easier.I think it was misprint that wasted a week of my time. thanks anyways
$endgroup$
– Stomp
2 days ago
1
1
$begingroup$
I don't understand what "$AG$ is produced to $X$" means. Did you draw a picture? Can you share it with us?
$endgroup$
– Matteo
2 days ago
$begingroup$
I don't understand what "$AG$ is produced to $X$" means. Did you draw a picture? Can you share it with us?
$endgroup$
– Matteo
2 days ago
3
3
$begingroup$
Are you sure that $GX = AC$? I've calculated that you need $GX=AG$ for this theorem to be true.
$endgroup$
– Adam Latosiński
2 days ago
$begingroup$
Are you sure that $GX = AC$? I've calculated that you need $GX=AG$ for this theorem to be true.
$endgroup$
– Adam Latosiński
2 days ago
$begingroup$
Could you provide a picture??
$endgroup$
– Dr. Mathva
2 days ago
$begingroup$
Could you provide a picture??
$endgroup$
– Dr. Mathva
2 days ago
2
2
$begingroup$
@Matteo "Produce" is a slightly old-fashioned term in geometry meaning to extend or lengthen. You can find the definition in Wiktionary.
$endgroup$
– FredH
2 days ago
$begingroup$
@Matteo "Produce" is a slightly old-fashioned term in geometry meaning to extend or lengthen. You can find the definition in Wiktionary.
$endgroup$
– FredH
2 days ago
$begingroup$
Thanks you guys ,if AG is equal to GX then the problem becomes much easier.I think it was misprint that wasted a week of my time. thanks anyways
$endgroup$
– Stomp
2 days ago
$begingroup$
Thanks you guys ,if AG is equal to GX then the problem becomes much easier.I think it was misprint that wasted a week of my time. thanks anyways
$endgroup$
– Stomp
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I can prove that you'd need $X$ such that $|GX| = |AG|$ using vector algebra.
Let us describe the points of the plane using vectors from point A. Then $vecA = vec0$, and $vecB$ and $vecC$ are linearly independent. The centroid $G$ is given by a vector
$$ vecG = frac13(vecA+vecB+vecC) = frac13(vecB+vecC) $$
Point $X$ lies on the line $overlineAG$, which means that
$$ exists lambdainmathbbR : vecX = lambda (vecB+vecC) $$
For point $L$ we have
beginalign big(Lin overlineBCbig) &Rightarrow big(exists alpha_LinmathbbR : vecL = alpha_L vecB + (1-alpha_L) vecC big) \
big(overlineXL parallel overlineAC big) &Rightarrow big(exists beta_LinmathbbR : vecL = vecX + beta_L vecC = lambda vecB + (lambda + beta_L)vecCbig) endalign
since vectors $vecB=vecC$ the only solution for these conditions is
$$ vecL = lambda vecB + (1-lambda)vecC$$
For point $M$ we have
beginalign big(Min overlineACbig) &Rightarrow big(exists alpha_MinmathbbR : vecM = alpha_M vecC big) \
big(overlineXMparallel overlineABbig) &Rightarrow big(exists beta_MinmathbbR : vecM = vecX + beta_M vecB = (lambda + beta_M) vecB + lambda vecCbig) endalign
The only solution for these conditions is
$$ vecM = lambda vecC$$
Finally, for point $N$ we have
beginalign big(Nin overlineABbig) &Rightarrow big(exists alpha_NinmathbbR : vecN = alpha_N vecB big) \
big(overlineXNparallel overlineBCbig) &Rightarrow big(exists beta_NinmathbbR : vecN = vecX + beta_N (vecB-vecC) = (lambda + beta_N) vecB + (lambda-beta_N) vecCbig) endalign
and the solution for these conditions is
$$ vecN = 2lambda vecB$$
Now, if $L$, $M$, $N$ are supposed to be colinear that means that
beginalign existsgammainmathbbR &: (vecL-vecM) = gamma (vecN-vecL) \
existsgammainmathbbR &: lambdavecB+(1-2lambda)vecC = gamma (lambdavecB+(lambda-1)vecC)endalign
This can only be true if $lambda=frac23$, or $lambda=0$. The second option would correspond to $X=A$ and it isn't the case we're interested in. That means that $vecX=frac23(vecB+vecC) = 2vecG$
If you choose any other point $X$ on line $overlineAG$, points $L$, $M$, $N$ won't be colinear.
answered 2 days ago
Adam LatosińskiAdam Latosiński
3836
$endgroup$
$begingroup$
Thank you, if AG=GX then the problem becomes solvable with similarity.Can't thank you enough for that.
$endgroup$
– Stomp
2 days ago
add a comment |
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$begingroup$
I can prove that you'd need $X$ such that $|GX| = |AG|$ using vector algebra.
Let us describe the points of the plane using vectors from point A. Then $vecA = vec0$, and $vecB$ and $vecC$ are linearly independent. The centroid $G$ is given by a vector
$$ vecG = frac13(vecA+vecB+vecC) = frac13(vecB+vecC) $$
Point $X$ lies on the line $overlineAG$, which means that
$$ exists lambdainmathbbR : vecX = lambda (vecB+vecC) $$
For point $L$ we have
beginalign big(Lin overlineBCbig) &Rightarrow big(exists alpha_LinmathbbR : vecL = alpha_L vecB + (1-alpha_L) vecC big) \
big(overlineXL parallel overlineAC big) &Rightarrow big(exists beta_LinmathbbR : vecL = vecX + beta_L vecC = lambda vecB + (lambda + beta_L)vecCbig) endalign
since vectors $vecB=vecC$ the only solution for these conditions is
$$ vecL = lambda vecB + (1-lambda)vecC$$
For point $M$ we have
beginalign big(Min overlineACbig) &Rightarrow big(exists alpha_MinmathbbR : vecM = alpha_M vecC big) \
big(overlineXMparallel overlineABbig) &Rightarrow big(exists beta_MinmathbbR : vecM = vecX + beta_M vecB = (lambda + beta_M) vecB + lambda vecCbig) endalign
The only solution for these conditions is
$$ vecM = lambda vecC$$
Finally, for point $N$ we have
beginalign big(Nin overlineABbig) &Rightarrow big(exists alpha_NinmathbbR : vecN = alpha_N vecB big) \
big(overlineXNparallel overlineBCbig) &Rightarrow big(exists beta_NinmathbbR : vecN = vecX + beta_N (vecB-vecC) = (lambda + beta_N) vecB + (lambda-beta_N) vecCbig) endalign
and the solution for these conditions is
$$ vecN = 2lambda vecB$$
Now, if $L$, $M$, $N$ are supposed to be colinear that means that
beginalign existsgammainmathbbR &: (vecL-vecM) = gamma (vecN-vecL) \
existsgammainmathbbR &: lambdavecB+(1-2lambda)vecC = gamma (lambdavecB+(lambda-1)vecC)endalign
This can only be true if $lambda=frac23$, or $lambda=0$. The second option would correspond to $X=A$ and it isn't the case we're interested in. That means that $vecX=frac23(vecB+vecC) = 2vecG$
If you choose any other point $X$ on line $overlineAG$, points $L$, $M$, $N$ won't be colinear.
answered 2 days ago
Adam LatosińskiAdam Latosiński
3836
$endgroup$
$begingroup$
Thank you, if AG=GX then the problem becomes solvable with similarity.Can't thank you enough for that.
$endgroup$
– Stomp
2 days ago
add a comment |
$begingroup$
I can prove that you'd need $X$ such that $|GX| = |AG|$ using vector algebra.
Let us describe the points of the plane using vectors from point A. Then $vecA = vec0$, and $vecB$ and $vecC$ are linearly independent. The centroid $G$ is given by a vector
$$ vecG = frac13(vecA+vecB+vecC) = frac13(vecB+vecC) $$
Point $X$ lies on the line $overlineAG$, which means that
$$ exists lambdainmathbbR : vecX = lambda (vecB+vecC) $$
For point $L$ we have
beginalign big(Lin overlineBCbig) &Rightarrow big(exists alpha_LinmathbbR : vecL = alpha_L vecB + (1-alpha_L) vecC big) \
big(overlineXL parallel overlineAC big) &Rightarrow big(exists beta_LinmathbbR : vecL = vecX + beta_L vecC = lambda vecB + (lambda + beta_L)vecCbig) endalign
since vectors $vecB=vecC$ the only solution for these conditions is
$$ vecL = lambda vecB + (1-lambda)vecC$$
For point $M$ we have
beginalign big(Min overlineACbig) &Rightarrow big(exists alpha_MinmathbbR : vecM = alpha_M vecC big) \
big(overlineXMparallel overlineABbig) &Rightarrow big(exists beta_MinmathbbR : vecM = vecX + beta_M vecB = (lambda + beta_M) vecB + lambda vecCbig) endalign
The only solution for these conditions is
$$ vecM = lambda vecC$$
Finally, for point $N$ we have
beginalign big(Nin overlineABbig) &Rightarrow big(exists alpha_NinmathbbR : vecN = alpha_N vecB big) \
big(overlineXNparallel overlineBCbig) &Rightarrow big(exists beta_NinmathbbR : vecN = vecX + beta_N (vecB-vecC) = (lambda + beta_N) vecB + (lambda-beta_N) vecCbig) endalign
and the solution for these conditions is
$$ vecN = 2lambda vecB$$
Now, if $L$, $M$, $N$ are supposed to be colinear that means that
beginalign existsgammainmathbbR &: (vecL-vecM) = gamma (vecN-vecL) \
existsgammainmathbbR &: lambdavecB+(1-2lambda)vecC = gamma (lambdavecB+(lambda-1)vecC)endalign
This can only be true if $lambda=frac23$, or $lambda=0$. The second option would correspond to $X=A$ and it isn't the case we're interested in. That means that $vecX=frac23(vecB+vecC) = 2vecG$
If you choose any other point $X$ on line $overlineAG$, points $L$, $M$, $N$ won't be colinear.
answered 2 days ago
Adam LatosińskiAdam Latosiński
3836
$endgroup$
$begingroup$
Thank you, if AG=GX then the problem becomes solvable with similarity.Can't thank you enough for that.
$endgroup$
– Stomp
2 days ago
add a comment |
$begingroup$
I can prove that you'd need $X$ such that $|GX| = |AG|$ using vector algebra.
Let us describe the points of the plane using vectors from point A. Then $vecA = vec0$, and $vecB$ and $vecC$ are linearly independent. The centroid $G$ is given by a vector
$$ vecG = frac13(vecA+vecB+vecC) = frac13(vecB+vecC) $$
Point $X$ lies on the line $overlineAG$, which means that
$$ exists lambdainmathbbR : vecX = lambda (vecB+vecC) $$
For point $L$ we have
beginalign big(Lin overlineBCbig) &Rightarrow big(exists alpha_LinmathbbR : vecL = alpha_L vecB + (1-alpha_L) vecC big) \
big(overlineXL parallel overlineAC big) &Rightarrow big(exists beta_LinmathbbR : vecL = vecX + beta_L vecC = lambda vecB + (lambda + beta_L)vecCbig) endalign
since vectors $vecB=vecC$ the only solution for these conditions is
$$ vecL = lambda vecB + (1-lambda)vecC$$
For point $M$ we have
beginalign big(Min overlineACbig) &Rightarrow big(exists alpha_MinmathbbR : vecM = alpha_M vecC big) \
big(overlineXMparallel overlineABbig) &Rightarrow big(exists beta_MinmathbbR : vecM = vecX + beta_M vecB = (lambda + beta_M) vecB + lambda vecCbig) endalign
The only solution for these conditions is
$$ vecM = lambda vecC$$
Finally, for point $N$ we have
beginalign big(Nin overlineABbig) &Rightarrow big(exists alpha_NinmathbbR : vecN = alpha_N vecB big) \
big(overlineXNparallel overlineBCbig) &Rightarrow big(exists beta_NinmathbbR : vecN = vecX + beta_N (vecB-vecC) = (lambda + beta_N) vecB + (lambda-beta_N) vecCbig) endalign
and the solution for these conditions is
$$ vecN = 2lambda vecB$$
Now, if $L$, $M$, $N$ are supposed to be colinear that means that
beginalign existsgammainmathbbR &: (vecL-vecM) = gamma (vecN-vecL) \
existsgammainmathbbR &: lambdavecB+(1-2lambda)vecC = gamma (lambdavecB+(lambda-1)vecC)endalign
This can only be true if $lambda=frac23$, or $lambda=0$. The second option would correspond to $X=A$ and it isn't the case we're interested in. That means that $vecX=frac23(vecB+vecC) = 2vecG$
If you choose any other point $X$ on line $overlineAG$, points $L$, $M$, $N$ won't be colinear.
answered 2 days ago
Adam LatosińskiAdam Latosiński
3836
$endgroup$
I can prove that you'd need $X$ such that $|GX| = |AG|$ using vector algebra.
Let us describe the points of the plane using vectors from point A. Then $vecA = vec0$, and $vecB$ and $vecC$ are linearly independent. The centroid $G$ is given by a vector
$$ vecG = frac13(vecA+vecB+vecC) = frac13(vecB+vecC) $$
Point $X$ lies on the line $overlineAG$, which means that
$$ exists lambdainmathbbR : vecX = lambda (vecB+vecC) $$
For point $L$ we have
beginalign big(Lin overlineBCbig) &Rightarrow big(exists alpha_LinmathbbR : vecL = alpha_L vecB + (1-alpha_L) vecC big) \
big(overlineXL parallel overlineAC big) &Rightarrow big(exists beta_LinmathbbR : vecL = vecX + beta_L vecC = lambda vecB + (lambda + beta_L)vecCbig) endalign
since vectors $vecB=vecC$ the only solution for these conditions is
$$ vecL = lambda vecB + (1-lambda)vecC$$
For point $M$ we have
beginalign big(Min overlineACbig) &Rightarrow big(exists alpha_MinmathbbR : vecM = alpha_M vecC big) \
big(overlineXMparallel overlineABbig) &Rightarrow big(exists beta_MinmathbbR : vecM = vecX + beta_M vecB = (lambda + beta_M) vecB + lambda vecCbig) endalign
The only solution for these conditions is
$$ vecM = lambda vecC$$
Finally, for point $N$ we have
beginalign big(Nin overlineABbig) &Rightarrow big(exists alpha_NinmathbbR : vecN = alpha_N vecB big) \
big(overlineXNparallel overlineBCbig) &Rightarrow big(exists beta_NinmathbbR : vecN = vecX + beta_N (vecB-vecC) = (lambda + beta_N) vecB + (lambda-beta_N) vecCbig) endalign
and the solution for these conditions is
$$ vecN = 2lambda vecB$$
Now, if $L$, $M$, $N$ are supposed to be colinear that means that
beginalign existsgammainmathbbR &: (vecL-vecM) = gamma (vecN-vecL) \
existsgammainmathbbR &: lambdavecB+(1-2lambda)vecC = gamma (lambdavecB+(lambda-1)vecC)endalign
This can only be true if $lambda=frac23$, or $lambda=0$. The second option would correspond to $X=A$ and it isn't the case we're interested in. That means that $vecX=frac23(vecB+vecC) = 2vecG$
If you choose any other point $X$ on line $overlineAG$, points $L$, $M$, $N$ won't be colinear.
answered 2 days ago
Adam LatosińskiAdam Latosiński
3836
answered 2 days ago
Adam LatosińskiAdam Latosiński
3836
answered 2 days ago
Adam LatosińskiAdam Latosiński
3836
answered 2 days ago
Adam LatosińskiAdam Latosiński
3836
3836
$begingroup$
Thank you, if AG=GX then the problem becomes solvable with similarity.Can't thank you enough for that.
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– Stomp
2 days ago
add a comment |
$begingroup$
Thank you, if AG=GX then the problem becomes solvable with similarity.Can't thank you enough for that.
$endgroup$
– Stomp
2 days ago
$begingroup$
Thank you, if AG=GX then the problem becomes solvable with similarity.Can't thank you enough for that.
$endgroup$
– Stomp
2 days ago
$begingroup$
Thank you, if AG=GX then the problem becomes solvable with similarity.Can't thank you enough for that.
$endgroup$
– Stomp
2 days ago
add a comment |
Stomp is a new contributor. Be nice, and check out our Code of Conduct.
Stomp is a new contributor. Be nice, and check out our Code of Conduct.
Stomp is a new contributor. Be nice, and check out our Code of Conduct.
Stomp is a new contributor. Be nice, and check out our Code of Conduct.
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1
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I don't understand what "$AG$ is produced to $X$" means. Did you draw a picture? Can you share it with us?
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– Matteo
2 days ago
3
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Are you sure that $GX = AC$? I've calculated that you need $GX=AG$ for this theorem to be true.
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– Adam Latosiński
2 days ago
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Could you provide a picture??
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– Dr. Mathva
2 days ago
2
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@Matteo "Produce" is a slightly old-fashioned term in geometry meaning to extend or lengthen. You can find the definition in Wiktionary.
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– FredH
2 days ago
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Thanks you guys ,if AG is equal to GX then the problem becomes much easier.I think it was misprint that wasted a week of my time. thanks anyways
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– Stomp
2 days ago