Question

Prove that the line segment 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 that:-

$△ABC$ in whichD, E, F are the mid points of sides AB, BC and CA respectively.

To prove:- each of the triangles are similar to the original triangle, i.e.,

$△ADF\sim △ABC$

$△BDE\sim △ABC$

$△CEF\sim △ABC$

Proof:-

Consider the $△ADF$ and $△ABC$

Since D and F are the mid points of AB and AC respectively.

$\therefore DF\parallel BC$

$\Rightarrow \angle AFD=\angle B\left(\text{Corresponding angles are equal}\right)$

Now, in $△ADF$ and $△ABC$, we have

$\angle ADF=\angle B\left(\text{Corresponding angles}\right)$

$\angle A=\angle A\left(\text{Common}\right)$

By AA similar conditions,

$△ADF\sim △ABC$

Similarly, we have

$△BDE\sim △ABC$

$△CEF\sim △ABC$

$\therefore EF\parallel AB$

$\Rightarrow EF\parallel AD..........\left(1\right)$

And, $DE\parallel AC$

$\Rightarrow DE\parallel AF..........\left(2\right)$

From ${eq}^n\left(1\right)\&\left(2\right)$, we have

ADEF is a parallelogram.

Similarly, BDFE is a parallelogram.

Now, in $△ABC$ and $△DEF$

$\angle A=\angle FED(\because \text{Opposite angles of parallelogram})$

$\angle B=\angle DFE(\because \text{Opposite angles of parallelogram})$

Therefore, by AA similar condition

$△ABC\sim △DEF$

Hence proved that each of the triangles are similar to the original triangle.