At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
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?
Let me read it just a sec
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 "
OK, I got it... right, you are right, what next?
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
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?
yes. And the proof relies on vectors, and the fact that the dot product of two vectors that are perpendicular = 0
What is the need of that constant l in the equation \(la.(c−b)=0\)
good question. It is obviously true, but irrelevant. I would just claim a dot (c-b)= 0
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.
but vectors do not have to "intersect"
I assume that relabeling OA as a, etc makes sense. Do you see how BC= c - b (where all 3 are treated as vectors)?
Personally, I always "think" vector addition BC+B= c (using head to tail to add), and then rearrange the vectors to get the difference.
*b (not B)
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 :)
Thanks a lot
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..