Let P,Q be mid-points of AB, AC in ∇ ABC

Draw AM parallel to BC such that MQN, N being in BC, is parallel to AB.

       ^AQM = ^NQC         (opp. angles of intersecting lines)

       ^QAM = ^QCN        (alt. angles in parallel lines)

      ^AMQ = ^QNC        (alt. angles in parallel lines)

              AQ = QC                (construct)

∴  ∇ AMQ = ∇ CNQ       (tri.s 2 ang.s + inc. side equal)

  ∴        QM = QN

ABNM is a parallelogram     (opp. Sides parallel see above)

∴            AB = MN             (opp. Sides parallelogram eq.)

∴            AP = MQ             (as AP = PB & MQ = QM)

       ^AQM = ^QAP         (alt. ang.s AB parallel MN)

               AQ = AQ

∴   ∇ AMQ = ∇ APQ        (2 sides & inc. ang. eq.)

∴      ^AQP = ^QAM

But 

       ^QAM = ^QCN         (alt. angles in parallel lines)

∴      ^AQP = ^QCN

∴  PQ is parallel to BC    (Corr. ang.s eq.)

Created 23rd July 2008