Triangle altitudes meet at a single point

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