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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 \( \triangle ABC \).

 

 Jan 20, 2021
 #1
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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   

 

 

cool cool cool

 Jan 20, 2021

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