XY is a line parallel to side BC of a ΔABC. If BE⊥AC and CF⊥AB meet XY at E and F, respectively. Show that ar (ΔABE) = ar (ΔACF).

Given :XY∥BCandCF∥AB⇒XF∥BCandCF∥AB⇒BCFXis a parallelogram.ar(△ACF)=12ar(BCFX)... (1) (△ACFand || gmBCFXare on the same baseCFand between the same parallelABandFC)BE∥CYandXY∥BC(given)So,BCYEis a parallelogram.ar(△ABE)=12ar(BCYE)... (2) (△ABEand ||gmBCYEare on the same baseBEand between the same parallel linesACandBE)ar(BCFX)=ar(BCYE)... (3) (ParallelogramBCFXandBCYEare on the same baseBCand between the same parallelsBCandEF.From (1) , (2) and (3) ,ar(△ABE)=ar(△ACF)