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## Answers

**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

HENCE PROVED