Question

In given figure prove that the line segments joining the mid-points of the sides of a triangle form four triangles, each of which is similar to the original triangle.

Answer 1

Given $△ABC$ in which D,E,F are the mid-points of sides BC,CA and AB respectively.

To prove Each of the triangles AFE, FBD, EDC and DEF is similar to $△ABC$.

Proof: Consider triangles AFE and ABC.

Since F and E are mid-points of AB and AC respectively

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$\therefore$ $FE||BC$

$\Rightarrow$ $\angle AFE=\angle B$ [Corresponding angles]

Thus, in $△AFE$ and $△ABC$, we have

$\angle AFE=\angle B$ and,

$\angle A=\angle A$ [Common]

$\therefore$ $△AFE\sim △ABC$

Similarly, we have

$△FBD\sim △ABC$ and $△EDC\sim △ABC$

Now, we shall show that $△DEF\sim △ABC$.

Clearly, ED||AF and DF||EA.

$\therefore$ AFDE is a parallelogram.

$\Rightarrow$ $\angle EDF=\angle A$ [$\because$ Oppositte angles of a parallelogram are equal]

Similarly, BDEF is a parallelogram.

$\therefore$ $\angle DEF=\angle B$ [$\because$ Opposite angles of a parallelogram are equal]

Thus, in triangles DEF and ABC, we have

$\angle EDF=\angle A$ and $\angle DEF=\angle B$

So, by AA-criterion of similarity, we have

$△DEF\sim △ABC$

Thus, each one of the triangles AFE, FBD, EDC and DEF is similar to $△ABC$