text: # Geometrical Proof
  
  ## Homework Statement
  
  Consider an triangle ABC with M as the middle point of the side AB.
  On the straight line through AB you put the angle ∠ ACM at A and the angle ∠ MCB at B. Now you have two new lines. The new lines should be on the same side of AB as C.
  Proof that the intersection point of the two new lines is located on the line through CM.
  
  ## The Attempt at a Solution
  
  I wanted to use Cevas Theorem but I could not use it :(
  
  I hope you can give me some advice.
  
  You've generated some similar triangles there which should help, if you work out the scaling. Note that |AM| = |BM|
  
  Mark44
  Mentor
  
  ## Homework Statement
  
  Consider an triangle ABC with M as the middle point of the side AB.
  On the straight line through AB you put the angle ∠ ACM at A and the angle ∠ MCB at B. Now you have two new lines.
  I don't understand. From your explanation you would have two new angles. In my drawing below I have labelled ∠ ACM as a and ∠ MCB as b.
  franceboy said:
  The new lines should be on the same side of AB as C.
  Proof that the intersection point of the two new lines is located on the line through CM.
  ???
  franceboy said:
  
  ## The Attempt at a Solution
  
  I wanted to use Cevas Theorem but I could not use it :(
  What is Ceva's Theorem?
  franceboy said:
  I hope you can give me some advice.
  
  Hi Mark, this is the construction as I understand it:
  
  Mark44
  Mentor
  Sorry thai i did not add a sketch but a GeoGebra sketch was not accepted as file.
  Joffan your sketch is correct. I used the symmetry so that I " only" need to proof
  BX / XC * CY / YA = 1
  I determined all the angles and I found some similarities but they did not help to solve the problem.
  Is Ceva' s theorem the right idea to solve the problem?
  
  Maybe you could use Ceva's theorem - it seems a little overpowered.
  
  I would proceed by marking X as the intersection of the new line from A with CM and Y as the intersection of the new line from B with CM. Then show that |MX| = |MY| and thus that X == Y
explanation: The document is a discussion about a geometrical proof problem. It contains a question about how to prove that the intersection point of two new lines is located on the line through CM in a given triangle configuration. The user has attempted to solve the problem using Ceva's Theorem but has encountered difficulties. The discussion includes feedback and suggestions from another user, Mark44, and further clarification from the original poster.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
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distilabel_metadata.role: system
distilabel_metadata.content: # Geometrical Proof
  
  ## Homework Statement
  
  Consider an triangle ABC with M as the middle point of the side AB.
  On the straight line through AB you put the angle ∠ ACM at A and the angle ∠ MCB at B. Now you have two new lines. The new lines should be on the same side of AB as C.
  Proof that the intersection point of the two new lines is located on the line through CM.
  
  ## The Attempt at a Solution
  
  I wanted to use Cevas Theorem but I could not use it :(
  
  I hope you can give me some advice.
  
  You've generated some similar triangles there which should help, if you work out the scaling. Note that |AM| = |BM|
  
  Mark44
  Mentor
  
  ## Homework Statement
  
  Consider an triangle ABC with M as the middle point of the side AB.
  On the straight line through AB you put the angle ∠ ACM at A and the angle ∠ MCB at B. Now you have two new lines.
  I don't understand. From your explanation you would have two new angles. In my drawing below I have labelled ∠ ACM as a and ∠ MCB as b.
  franceboy said:
  The new lines should be on the same side of AB as C.
  Proof that the intersection point of the two new lines is located on the line through CM.
  ???
  franceboy said:
  
  ## The Attempt at a Solution
  
  I wanted to use Cevas Theorem but I could not use it :(
  What is Ceva's Theorem?
  franceboy said:
  I hope you can give me some advice.
  
  Hi Mark, this is the construction as I understand it:
  
  Mark44
  Mentor
  Sorry thai i did not add a sketch but a GeoGebra sketch was not accepted as file.
  Joffan your sketch is correct. I used the symmetry so that I " only" need to proof
  BX / XC * CY / YA = 1
  I determined all the angles and I found some similarities but they did not help to solve the problem.
  Is Ceva' s theorem the right idea to solve the problem?
  
  Maybe you could use Ceva's theorem - it seems a little overpowered.
  
  I would proceed by marking X as the intersection of the new line from A with CM and Y as the intersection of the new line from B with CM. Then show that |MX| = |MY| and thus that X == Y
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document is a discussion about a geometrical proof problem. It contains a question about how to prove that the intersection point of two new lines is located on the line through CM in a given triangle configuration. The user has attempted to solve the problem using Ceva's Theorem but has encountered difficulties. The discussion includes feedback and suggestions from another user, Mark44, and further clarification from the original poster.
  
  Label: question-and-answer