(Euclid Book 1 Prop 34 - Part)

Construct parallelogram ABCD and join BD

Consider triangles ABD & CDB

      AB = CD                    (Opp. sides of parallelogram) 

      AD = CB .                  (Opp. sides of parallelogram)

^BAD = ^DCB             (Opp. angles of parallelogram) 

So   

∇ ABD = ∇CDB           ( Two sides and inc. angle equal)

∴  area ∇ABD = 1/2 area parallelogram ABCD

Similarly it can be shown ∇ACD = ∇BAC and area ∇ACD = 1/2 area parallelogram ABCD