Prove that the altitudes of a triangle are concurrent

[Martoman]

So i'm bored and want to determine if the altitudes of a triangle are indeed concurrent by vector methods.

I have http://img685.imageshack.us/i/prooconf.jpg/

Half of the problem was getting it into this form and declaring those vectors.

Now from that diagram i know that
is perpendicular to AC.


Hence

Similarly for the other side BC

Hrmmm, now I am wondering where to go from here.

I was thinking along the lines of

Then
and similarly

Then they all intersect at the same point, but I think this reasoning is fallacious as it assumes the angles to be the same.... correct?

If this doesn't work, then where to go?

[kamil9876]

correct: you can't assume the are equal.

A good way to finish it is to stretch out the line segment so that it hits . This is vector is basically there is a non-zero real number (scale factor).

Now you wish to show is perpendicular to , so try to do a dot product again.


Great start on the proof btw, I remember comming up with something similair for this one:

"Prove that if the midpoint of each line segment is connected to the point on the opposite side, then these three line segments are concurrent"

[Martoman]

Yes that was where I started as well with that proof "prove that the medians of a triangle are concurrent". That was bitchy, but a bit easier than this one.

[Martoman]

correct: you can't assume the are equal.

A good way to finish it is to stretch out the line segment so that it hits . This is vector is basically there is a non-zero real number (scale factor).

Now you wish to show is perpendicular to , so try to do a dot product again.


Great start on the proof btw, I remember comming up with something similair for this one:

"Prove that if the midpoint of each line segment is connected to the point on the opposite side, then these three line segments are concurrent"

I may be being silly here.... given its 1 in the freaking morning... but if perp to AB means Then the lambda just gets canceled out.

[kamil9876]

yeah true, just makes the expression easier. I pointed out is non-zero, and that is why you can 'cancel' it.

[Martoman]

Oh. Disgusting. Interestingly, i just thought if you inscribe it in a circle then instantly |a|=|b|=|c| but it isn't a vector proof. Fuck my creativity.