In \( \triangle ABC \), we place points \( D\), \(E\), and \(F \) on sides \( AB\), \(BC\), and \(CA \) respectively, such that \( \dfrac{\overline{AD}}{\overline{AB}} =\dfrac{\overline{BE}}{\overline{BC}}=\dfrac{\overline{CF}}{\overline{CA}} = \dfrac{1}{3} \). Find the ratio of the area of \( \triangle DEF \) to the area of \( \triangle ABC \).
+128473 The area of ABC = (1/2)(AC)(BC) sin ACB = (AC))BC) sin ABC / 2
And the area of triangle FCE = (1/2) (AC/3) ( 2BC/3) sin ACB = (AC) (BC) sin ACB / 9
So [ FCE ] / [ ABC ] = 2/9 ⇒ [FCE] = (2/9) [ ABC]
And each of the white areas = 2/9 area of ABC
So their total area = 3(2/9)area of ABC = 2/3 area of ABC
So area of DEF / area of ABC = [ area of ABC - 2/3 area of ABC ] / area of ABC = 1/3
