Euler Line theoram - part of the proof is not clear to me

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    Euler Line Proof
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The discussion revolves around the proof of the Euler Line theorem, specifically regarding the collinearity of the circumcenter, centroid, and orthocenter of a triangle. A participant expresses confusion about a step in the proof that concludes CK is parallel to OD, which would imply that CH* and OD are also parallel. They acknowledge the similarity of triangles DOG and CH*G but question the validity of concluding parallel lines from equal angles. The conversation highlights a need for clarification on the theorem that states equal angles imply parallel lines, as well as the distinction between similar triangles and parallel sides. The discussion ultimately seeks to resolve the misunderstanding surrounding the parallelism in the proof.
musicgold
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This is not homework. I am reading a book: "The art of infinite: The Pleasure of Mathematics" and pages 119-120 give a proof of the Euler Line theoram: the circumcenter, centroild and orthocenter of a triangle are always colinear (see the attached files).

1. Homework Statement

Page 119 shows a triangle with tree points. O, G, and H* are colinear. O is the circumcenter and G is the centroid of the triangle. H* is a point we hope to prove to be the orthocenter (H) of the triangle.

I am not clear on one point in the proof. I have put a question mark against the underlined part on page 120.
I am not sure how we can concldue that CK is parallel to OD (and therefore parpendicular to AB).

Homework Equations

The Attempt at a Solution


While I can see that ## \Delta## DOG and ## \Delta##CH*G are similar, but I am not sure how we can jump to the concusion that CH* and OD are parallel. What am I missing?
 

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Hi,
If you agree that the two triangles are similar, ##\angle CH^* G = \angle DOG## and there is a theorem: angles equal ##\Leftrightarrow## lines parallel
 
BvU said:
Hi,
there is a theorem: angles equal ##\Leftrightarrow## lines parallel
Thanks.
Could you please point me to that theorem?

I know that if we start with two parallel lines and add two transversal lines, we get two similar triangles. However, here we are concluding that the lines are parallel. Can we not have have two similar triangles where neither of the sides are parallel?

For example, the solution of problem 4 on the following page has two similar triangles but none of the lines are parallel. What am I missing?
 
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I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks

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