Let ABC be any triangle; ABDE, BCFG, and CAHI parallelograms constructed externally on the sides of ABC; and P the intersection of lines FG and HI.

If AECP is a parallelogram, then prove that

area(ABDE) = area(BCFG) + area(CAHI).

Let J be the intersection of DB and PG.
Let K be the intersection of EA and PH.
Let M and N be the intersections of PC with AB and DE respectively.

EA = CP = DB  (opposite sides of parallelograms), therefore:
[ABDE] = [DNMB] + [NEAM]       using [ ] to denote area
            = [BCPJ] + [CAKP]         parallelograms with equal bases and heights
            = [BCFG] + [CAHI]         parallelograms with equal bases and heights

Posted by Harry on 2011-03-01 14:10:36