STATEMENT:IF A LINE IS DRAWN PARALLEL TO SIDE OF A TRIANGLE TO INTERSECT THE OTHER TWO SIDES IN DISTINCT POINTS THEN THE OTHER TWO SIDES ARE DIVIDE IN THE SAME RATIO.
GIVEN:In Δ ABC DE is parallel to BC which intersects sides AB and AC at D and E respectively
RTP: AD / DB = AE / EC
CONSTRUCTION:Join B,E and C,D and then draw DM is perpendicular to AC and EN perpendicular to AB
PROOF: Area of ΔADB = 1/2*AD*EN
AREA of ΔBDE = 1/2*BD*EN
so, ar(ΔADB) / ar(ΔBDE) = 1/2*AD*EN / 1/2*BD*EN = AD/BD -------(1)
area of ΔADE = 1/2*AE*DM
area of ΔCDE = 1/2*EN*DM
SO,ar(ΔADE) / ar(ΔCDE) = 1/2*AE*DM / 1/2*EN*DM = AE/EN ----------(2)
we observe that ΔBDE and ΔCDE are on the same base DE and between the same parallel BC and DE.
so ar(ΔBDE) = ar(ΔCDE) -------(3)
from (1),(2),(3) we know
AD/DB = AE/EC