# 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.

$$\therefore$$  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