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2bornot2b

  • 3 years ago

I am trying to solve the following problem, and I have a solution which I don't understand, can you help me? "Show that in a triangle the perpendiculars drawn from the vertices are concurrent. "

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  1. 2bornot2b
    • 3 years ago
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    Here is the solution that I don't understand

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  2. phi
    • 3 years ago
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    We start with a given: 2 of the perpendiculars (AD and BE) meet at point O now show that the third perpendicular also meets at point O i.e. if the line segment CO extended to side AB (at point F) is perpendicular to AB then all three perpendiculars meet at O Does that make sense?

  3. 2bornot2b
    • 3 years ago
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    Let me read it just a sec

  4. 2bornot2b
    • 3 years ago
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    This line can you state it in a better way, I don't get it "if the line segment CO extended to side AB (at point F) is perpendicular to AB then all three perpendiculars meet at O "

  5. 2bornot2b
    • 3 years ago
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    OK, I got it... right, you are right, what next?

  6. phi
    • 3 years ago
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    The 3rd perpendicular starts at vertex C. We can draw a line CO (C to O). if we continue, CO intersects the third side AB at F. Now if it turns out that COF is perpendicular to AB, then it is the perpendicular, and it goes through O, the same point as the other 2 perpendiculars

  7. 2bornot2b
    • 3 years ago
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    So we have to show that if the line CO is extended, and at the point where it meets AB it turns out to be perpendicular, then we are done right?

  8. phi
    • 3 years ago
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    yes. And the proof relies on vectors, and the fact that the dot product of two vectors that are perpendicular = 0

  9. 2bornot2b
    • 3 years ago
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    What is the need of that constant l in the equation \(la.(c−b)=0\)

  10. phi
    • 3 years ago
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    good question. It is obviously true, but irrelevant. I would just claim a dot (c-b)= 0

  11. phi
    • 3 years ago
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    Maybe they want to say that AO is too short (i.e. it does not reach the other side), but that we can scale it so that it does.

  12. phi
    • 3 years ago
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    but vectors do not have to "intersect"

  13. 2bornot2b
    • 3 years ago
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    OK

  14. phi
    • 3 years ago
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    I assume that relabeling OA as a, etc makes sense. Do you see how BC= c - b (where all 3 are treated as vectors)?

  15. phi
    • 3 years ago
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    Personally, I always "think" vector addition BC+B= c (using head to tail to add), and then rearrange the vectors to get the difference.

  16. phi
    • 3 years ago
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    *b (not B)

  17. 2bornot2b
    • 3 years ago
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    OK, so I will be going through the solution, and try to understand it, and if I find any problem, I will post it here. Please come again if you find a new post made on this problem :)

  18. phi
    • 3 years ago
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    roger that.

  19. 2bornot2b
    • 3 years ago
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    Thanks a lot

  20. 2bornot2b
    • 3 years ago
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    OK, its clear thank you! I wish I could provide you more medals. It's rare that someone digs in the unanswered questions, like you..

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