Show that parallell line of triangle is concurrent

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I found a rather interesting exercise which I have some problems with.

A line parallel with $AB$ and the triangle $Delta ABC $ intercepts the sides $AC$ and $BC$ at the points $A_1$, $B_1$.

Show that the lines $A_1B$, $AB_1$ and the median from $C$ is concurrent. In other words, show that the lines intersect at a single point.

Line of thought: Apply Cevas Theorem
If $A1,B1,C1$ are points on $AC$, $BC$ and $AB$ such that beginequationfracA C_1C_1Bcdot fracB B_1B_1 C cdot fracCA_1A_1A=1 quad (1)endequation then $AB_1$, $BA_1$ and $CC_1$ are concurrent.

The median from $C$ divides the side $AB$ in half such that $AC_1=C_1B$. Therefore $fracAC_1C_1B=1$.
$(1)$ now becomes beginequation fracBB_1B_1Ccdot fracCA_1A_1A=1 quad (2) endequation

Maybe we could use Thael's theorem triangle tells us that the ratio of any two corresponding sides of an equiangular triangle is always irrespective of their sizes. Thaels would give us that $$fracBB_1B_1C=fracA_1ACA_1$$ and if we insert this in $(2)$ the equality is correct.

This is my attempt but I'm pretty sure this is wrong. Seems like I'm doing some circular reasoning in the end. Any suggestions/help or even a completely other solution would be greatly appreciated!

geometry euclidean-geometry triangle

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edited Aug 2 at 17:13

Michael Rozenberg

87.2k1577178

asked Aug 2 at 16:37

einar

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up vote
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I found a rather interesting exercise which I have some problems with.

A line parallel with $AB$ and the triangle $Delta ABC $ intercepts the sides $AC$ and $BC$ at the points $A_1$, $B_1$.

Show that the lines $A_1B$, $AB_1$ and the median from $C$ is concurrent. In other words, show that the lines intersect at a single point.

Line of thought: Apply Cevas Theorem
If $A1,B1,C1$ are points on $AC$, $BC$ and $AB$ such that beginequationfracA C_1C_1Bcdot fracB B_1B_1 C cdot fracCA_1A_1A=1 quad (1)endequation then $AB_1$, $BA_1$ and $CC_1$ are concurrent.

The median from $C$ divides the side $AB$ in half such that $AC_1=C_1B$. Therefore $fracAC_1C_1B=1$.
$(1)$ now becomes beginequation fracBB_1B_1Ccdot fracCA_1A_1A=1 quad (2) endequation

Maybe we could use Thael's theorem triangle tells us that the ratio of any two corresponding sides of an equiangular triangle is always irrespective of their sizes. Thaels would give us that $$fracBB_1B_1C=fracA_1ACA_1$$ and if we insert this in $(2)$ the equality is correct.

This is my attempt but I'm pretty sure this is wrong. Seems like I'm doing some circular reasoning in the end. Any suggestions/help or even a completely other solution would be greatly appreciated!

geometry euclidean-geometry triangle

share|cite|improve this question

edited Aug 2 at 17:13

Michael Rozenberg

87.2k1577178

asked Aug 2 at 16:37

einar

1238

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

I found a rather interesting exercise which I have some problems with.

A line parallel with $AB$ and the triangle $Delta ABC $ intercepts the sides $AC$ and $BC$ at the points $A_1$, $B_1$.

Show that the lines $A_1B$, $AB_1$ and the median from $C$ is concurrent. In other words, show that the lines intersect at a single point.

Line of thought: Apply Cevas Theorem
If $A1,B1,C1$ are points on $AC$, $BC$ and $AB$ such that beginequationfracA C_1C_1Bcdot fracB B_1B_1 C cdot fracCA_1A_1A=1 quad (1)endequation then $AB_1$, $BA_1$ and $CC_1$ are concurrent.

The median from $C$ divides the side $AB$ in half such that $AC_1=C_1B$. Therefore $fracAC_1C_1B=1$.
$(1)$ now becomes beginequation fracBB_1B_1Ccdot fracCA_1A_1A=1 quad (2) endequation

Maybe we could use Thael's theorem triangle tells us that the ratio of any two corresponding sides of an equiangular triangle is always irrespective of their sizes. Thaels would give us that $$fracBB_1B_1C=fracA_1ACA_1$$ and if we insert this in $(2)$ the equality is correct.

This is my attempt but I'm pretty sure this is wrong. Seems like I'm doing some circular reasoning in the end. Any suggestions/help or even a completely other solution would be greatly appreciated!

geometry euclidean-geometry triangle

share|cite|improve this question

edited Aug 2 at 17:13

Michael Rozenberg

87.2k1577178

asked Aug 2 at 16:37

einar

1238

I found a rather interesting exercise which I have some problems with.

A line parallel with $AB$ and the triangle $Delta ABC $ intercepts the sides $AC$ and $BC$ at the points $A_1$, $B_1$.

Show that the lines $A_1B$, $AB_1$ and the median from $C$ is concurrent. In other words, show that the lines intersect at a single point.

Line of thought: Apply Cevas Theorem
If $A1,B1,C1$ are points on $AC$, $BC$ and $AB$ such that beginequationfracA C_1C_1Bcdot fracB B_1B_1 C cdot fracCA_1A_1A=1 quad (1)endequation then $AB_1$, $BA_1$ and $CC_1$ are concurrent.

The median from $C$ divides the side $AB$ in half such that $AC_1=C_1B$. Therefore $fracAC_1C_1B=1$.
$(1)$ now becomes beginequation fracBB_1B_1Ccdot fracCA_1A_1A=1 quad (2) endequation

Maybe we could use Thael's theorem triangle tells us that the ratio of any two corresponding sides of an equiangular triangle is always irrespective of their sizes. Thaels would give us that $$fracBB_1B_1C=fracA_1ACA_1$$ and if we insert this in $(2)$ the equality is correct.

This is my attempt but I'm pretty sure this is wrong. Seems like I'm doing some circular reasoning in the end. Any suggestions/help or even a completely other solution would be greatly appreciated!

geometry euclidean-geometry triangle

share|cite|improve this question

edited Aug 2 at 17:13

Michael Rozenberg

87.2k1577178

asked Aug 2 at 16:37

einar

1238

share|cite|improve this question

share|cite|improve this question

edited Aug 2 at 17:13

Michael Rozenberg

87.2k1577178

edited Aug 2 at 17:13

Michael Rozenberg

87.2k1577178

edited Aug 2 at 17:13

Michael Rozenberg

87.2k1577178

87.2k1577178

asked Aug 2 at 16:37

einar

1238

asked Aug 2 at 16:37

einar

1238

asked Aug 2 at 16:37

einar

1238

1238

add a comment | 

add a comment | 

1 Answer
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Let $AB_1cap BA_1=D$, $CDcap A_1B_1=E$ and $CDcap AB=F,$

$A_1E=a$, $EB_1=b$, $AF=c$ and $FB=d$.

Thus,
$$fracac=fracCECF=fracbd$$ and
$$fracad=fracEDDF=fracbc,$$ which gives $a=b$, $c=d$ and we are done!

share|cite|improve this answer

answered Aug 2 at 17:10

Michael Rozenberg

87.2k1577178

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank's for the answer! I don't see how CD∩AB? Could you possibly try to expand on this answer?
  – einar
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @einar In your case line $CD$ and line $AB$ have common point $F$. It's just the point $C_1$.
  – Michael Rozenberg
  6 hours ago
  

  
  
  
  

add a comment | 

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1 Answer
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1 Answer
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Let $AB_1cap BA_1=D$, $CDcap A_1B_1=E$ and $CDcap AB=F,$

$A_1E=a$, $EB_1=b$, $AF=c$ and $FB=d$.

Thus,
$$fracac=fracCECF=fracbd$$ and
$$fracad=fracEDDF=fracbc,$$ which gives $a=b$, $c=d$ and we are done!

share|cite|improve this answer

answered Aug 2 at 17:10

Michael Rozenberg

87.2k1577178

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank's for the answer! I don't see how CD∩AB? Could you possibly try to expand on this answer?
  – einar
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @einar In your case line $CD$ and line $AB$ have common point $F$. It's just the point $C_1$.
  – Michael Rozenberg
  6 hours ago
  

  
  
  
  

add a comment | 

up vote
1
down vote

Let $AB_1cap BA_1=D$, $CDcap A_1B_1=E$ and $CDcap AB=F,$

$A_1E=a$, $EB_1=b$, $AF=c$ and $FB=d$.

Thus,
$$fracac=fracCECF=fracbd$$ and
$$fracad=fracEDDF=fracbc,$$ which gives $a=b$, $c=d$ and we are done!

share|cite|improve this answer

answered Aug 2 at 17:10

Michael Rozenberg

87.2k1577178

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank's for the answer! I don't see how CD∩AB? Could you possibly try to expand on this answer?
  – einar
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @einar In your case line $CD$ and line $AB$ have common point $F$. It's just the point $C_1$.
  – Michael Rozenberg
  6 hours ago
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

Let $AB_1cap BA_1=D$, $CDcap A_1B_1=E$ and $CDcap AB=F,$

$A_1E=a$, $EB_1=b$, $AF=c$ and $FB=d$.

Thus,
$$fracac=fracCECF=fracbd$$ and
$$fracad=fracEDDF=fracbc,$$ which gives $a=b$, $c=d$ and we are done!

share|cite|improve this answer

answered Aug 2 at 17:10

Michael Rozenberg

87.2k1577178

Let $AB_1cap BA_1=D$, $CDcap A_1B_1=E$ and $CDcap AB=F,$

$A_1E=a$, $EB_1=b$, $AF=c$ and $FB=d$.

Thus,
$$fracac=fracCECF=fracbd$$ and
$$fracad=fracEDDF=fracbc,$$ which gives $a=b$, $c=d$ and we are done!

share|cite|improve this answer

answered Aug 2 at 17:10

Michael Rozenberg

87.2k1577178

share|cite|improve this answer

share|cite|improve this answer

answered Aug 2 at 17:10

Michael Rozenberg

87.2k1577178

answered Aug 2 at 17:10

Michael Rozenberg

87.2k1577178

answered Aug 2 at 17:10

Michael Rozenberg

87.2k1577178

87.2k1577178

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank's for the answer! I don't see how CD∩AB? Could you possibly try to expand on this answer?
  – einar
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @einar In your case line $CD$ and line $AB$ have common point $F$. It's just the point $C_1$.
  – Michael Rozenberg
  6 hours ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank's for the answer! I don't see how CD∩AB? Could you possibly try to expand on this answer?
  – einar
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @einar In your case line $CD$ and line $AB$ have common point $F$. It's just the point $C_1$.
  – Michael Rozenberg
  6 hours ago
  

  
  
  
  

Thank's for the answer! I don't see how CD∩AB? Could you possibly try to expand on this answer?
– einar
8 hours ago

Thank's for the answer! I don't see how CD∩AB? Could you possibly try to expand on this answer?
– einar
8 hours ago

@einar In your case line $CD$ and line $AB$ have common point $F$. It's just the point $C_1$.
– Michael Rozenberg
6 hours ago

@einar In your case line $CD$ and line $AB$ have common point $F$. It's just the point $C_1$.
– Michael Rozenberg
6 hours ago

add a comment | 

 

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