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2bornot2b
Group Title
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. "
 2 years ago
 2 years ago
2bornot2b Group Title
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. "
 2 years ago
 2 years ago

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2bornot2b Group TitleBest ResponseYou've already chosen the best response.0
Here is the solution that I don't understand
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
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?
 2 years ago

2bornot2b Group TitleBest ResponseYou've already chosen the best response.0
Let me read it just a sec
 2 years ago

2bornot2b Group TitleBest ResponseYou've already chosen the best response.0
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 "
 2 years ago

2bornot2b Group TitleBest ResponseYou've already chosen the best response.0
OK, I got it... right, you are right, what next?
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
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
 2 years ago

2bornot2b Group TitleBest ResponseYou've already chosen the best response.0
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?
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
yes. And the proof relies on vectors, and the fact that the dot product of two vectors that are perpendicular = 0
 2 years ago

2bornot2b Group TitleBest ResponseYou've already chosen the best response.0
What is the need of that constant l in the equation \(la.(c−b)=0\)
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
good question. It is obviously true, but irrelevant. I would just claim a dot (cb)= 0
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
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.
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
but vectors do not have to "intersect"
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
I assume that relabeling OA as a, etc makes sense. Do you see how BC= c  b (where all 3 are treated as vectors)?
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
Personally, I always "think" vector addition BC+B= c (using head to tail to add), and then rearrange the vectors to get the difference.
 2 years ago

2bornot2b Group TitleBest ResponseYou've already chosen the best response.0
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 :)
 2 years ago

2bornot2b Group TitleBest ResponseYou've already chosen the best response.0
Thanks a lot
 2 years ago

2bornot2b Group TitleBest ResponseYou've already chosen the best response.0
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..
 2 years ago
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