In a triangle ABC, DE parallel BC if AD: DB 3.5, then find ar(triangleDEF)÷ar(triangle CFB)
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F should be replaced by A bcoz there\’s no F in Δ
ΔABC~ΔADE….. (by SAS similarity)……… Now by theorem 6.6, ar(ΔABC)/ar(ΔADE)=(AB/AD)^2………..So ar(ΔABC)/ar(ΔADE)=(3/5)^2 = 9/25…