2bornot2b
  • 2bornot2b
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. "
Mathematics
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SOLVED
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jamiebookeater
  • jamiebookeater
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2bornot2b
  • 2bornot2b
Here is the solution that I don't understand
1 Attachment
phi
  • phi
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?
2bornot2b
  • 2bornot2b
Let me read it just a sec

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2bornot2b
  • 2bornot2b
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 "
2bornot2b
  • 2bornot2b
OK, I got it... right, you are right, what next?
phi
  • phi
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
2bornot2b
  • 2bornot2b
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?
phi
  • phi
yes. And the proof relies on vectors, and the fact that the dot product of two vectors that are perpendicular = 0
2bornot2b
  • 2bornot2b
What is the need of that constant l in the equation \(la.(c−b)=0\)
phi
  • phi
good question. It is obviously true, but irrelevant. I would just claim a dot (c-b)= 0
phi
  • phi
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.
phi
  • phi
but vectors do not have to "intersect"
2bornot2b
  • 2bornot2b
OK
phi
  • phi
I assume that relabeling OA as a, etc makes sense. Do you see how BC= c - b (where all 3 are treated as vectors)?
phi
  • phi
Personally, I always "think" vector addition BC+B= c (using head to tail to add), and then rearrange the vectors to get the difference.
phi
  • phi
*b (not B)
2bornot2b
  • 2bornot2b
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 :)
phi
  • phi
roger that.
2bornot2b
  • 2bornot2b
Thanks a lot
2bornot2b
  • 2bornot2b
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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