D, E, F are the mid-points of the sides BC, CA and AB respectively of a \(\triangle\) ABC. Determine the ratio of the areas of \(\triangle\) DEF and \(\triangle\) ABC.

Solution :

Since D and E are the mid-points of the sides BC and CA respectively of \(\triangle\) ABC.triangle

\(\therefore\)  DE || BA

\(\implies\)  DE || FA           ……….(1)

Since D and F are the mid-points of the sides BC and AB respectively of \(\triangle\) ABC. Therefore,

DF || CA  \(\implies\)  DF || AE         ………..(2)

From (1) and (2), we get AFDE is a parallelogram

Similarly, BDEF is a parallelogram.

In triangle DEF and ABC,

\(\angle\) FDE = \(\angle\) A          (opposite angles of ||gm AFDE)

\(\angle\) DEF = \(\angle\) B          (opposite angles of ||gm BDEF)

\(\therefore\)  By AA similarity,

\(\triangle\) DEF ~ \(\triangle\) ABC

\(\implies\)  \(area(\triangle DEF)\over area(\triangle ABC)\) = \({DE}^2\over {AB}^2\) = \(1\over 4\)

(The areas of two similar triangles are in the ratio of the squares of the corresponding sides)

Hence, Area of triangle DEF : Area of triangle ABC = 1 : 4

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