Proving orthocentre's existence

[Tahmid Hasan]

another i proved with intersecting chord theorem and some brute force transformations.try to figure it out.that makes 5 solutions.

[photon]

what's "brute force tranformations"???new words to me...

we know that the three perps bisect the angles of the orthic triangle and we also know that angle bisectors concur at one point(incenter).so the incenter of the orthic triangle in the orthocenter of the main triangle.

by the way,it was cool........

[Tahmid Hasan]

let $BE,CF$ be the perps which intersects at $H$.
connect $A,D$ and extend such that it intersects $BC$ at $D$.
so our task is to prove that $AD$ is also an altitude.
$B,C,E,F$ are concyclic($\angle BFC=\angle BEC$).
so $BH.HE=CH.HF$.
now let's draw a circle with diameter $AC$ it intersects $AD$ at $K$(extended if necessary).(btw i'm using contradiction here.)
so $\angle AKC=90$ and$AH.HK=HF.HC$.
hence$BH.HE=AH.HK$.
so $A,B,K,E$ is concyclic.
so $\angle AKB=90$.that's why $\angle AKC +\angle AKB = \pi$.
then $B,K,C$ are collinear.hence $K$ coincides with $D$.
so $AD$ is an altitude as we wished to prove