Prove that the altitudes of a triangle are concurrent.

Explanation

Let segments AD and CF be the altitudes of `triangleABC`, meeting each other in the point H.
then it is enough to prove that `bar(HB)` is perpen-
dicular to `bar(AC)`.
Choose H as the origin and let `bar(a),bar(b),bar(c)` be the position vectors of the vertices A, B and C respectively w.r.t the origin H.
Then `bar(HA)=bar(a),bar(HB)=bar(b)andbar(HC)=bar(c)`,
`bar(AB)=bar(b)-bar(a),bar(BC)=bar(c)andbar(AC)=bar(c)-bar(a)`.
Now `bar(HA)` is perpendicular to `bar(BC)`.
`:.bar(HA)*bar(BC)=0" ":.bar(a)*(bar(c)-bar(b))=0`
`:.bar(a)*bar(c)-bar(a)*bar(b)=0`
Alsho `bar(HC)` is perpendicular to `bar(AB)`.
`:.bar(HC)*bar(AB)=0" ":.bar(c)*(bar(a)-bar(a))=0`
`:.bar(c)*bar(b)-bar(c)*bar(a)=0`
`:.bar(c)*bar(b)-bar(a)*bar(c)=0" " . . .[because bar(c)*bar(a)=bar(a)*bar(c)]` . . .(2)
Adding (1) and (2), we get,
`bar(c)*bar(b)-bar(a)*bar(b)=0`
`:.(bar(c)-bar(a))*bar(b)=0`
`:.bar(b)*(bar(c)-bar(a))=0`
`:.bar(b)*(bar(c)-bar(a))=0`
`:.bar(HB)*bar(AC)=0`
`:.bar(HB)` is perpendicular to `bar(AC)`.
`:.` the altitudes of `triangleABD` are concurrent.