The three altitudes of any triangle (the perpendiculars from vertices to their opposite sides) intersect in a point called the orthocentre.
Consider any triangle ABC.
Draw AF perpendicular to CB, extend CB if necessary.
Draw BG perpendicular to CA, extend CA if necessary.
Draw CH perpendicular to BA, extend BA if necessary.
Draw PAQ parallel to BC.
Draw QCR parallel to AB.
Draw PBR parallel to AC
where P,Q,R are points of intersection.
PACB, BAQC are parallelograms (opp. sides parallel)
∴ PA=BC=AQ (opp. sides paral-gram equal)
∠AFC = 1 rt. ang. (construct)
= ∠FAP (alt. angs. BC||PQ
∴ AF is the perpendicular bisector of PQ in the larger ∇PQR
Similarly BPAC and RBAC are parallelograms and BG is the perpendicular bisector of PR in the larger ∇PQR
Similarly RBAC and BAQC are parallelograms and CH is the perpendicular bisector of PR in the larger triangle PQR
AF,BG,CH (extended if necessary) will all pass through a single point of intersection as they are perpendicular bisectors of the larger ∇PQR
Created 23rd July 2008