Geometrical Proof

  • Thread starter franceboy
  • Start date
  • #1
51
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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.

Homework Equations




The Attempt at a Solution


I wanted to use Ceva`s Theorem but I could not use it :(

I hope you can give me some advice.
 

Answers and Replies

  • #2
473
13
You've generated some similar triangles there which should help, if you work out the scaling. Note that |AM| = |BM|
 
  • #3
34,687
6,394

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:

Homework Equations




The Attempt at a Solution


I wanted to use Ceva`s Theorem but I could not use it :(
What is Ceva's Theorem?
franceboy said:
I hope you can give me some advice.
Triangle.jpg
 
  • #4
473
13
Hi Mark, this is the construction as I understand it:
PF_1.png
 
  • #6
51
0
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?
 
  • #7
473
13
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
 

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