Question

Using properties of similar triangles, prove that the line segment joining the mid-points of two sides of a triangle is:

(i) parallel to the third side.

(ii) one-half of the third side.

Answer 1

Step 1: Given data and draw a diagram for the situation:

Given: Line segment joins the mid-points of two sides of a triangle.

The diagrammatic representation is,

Step 2: Prove that the line segment joining the mid-points of two sides of a triangle is parallel to the third side:

It is given that, $A$ is mid-point of $PQ$ and $B$ is mid-point of $PR$.

$\Rightarrow PA=AQ$ and $\Rightarrow PB=BR$

In $△PQR$ and $△PAB$,

$\angle QPR=\angle APB$ …..$\left(\mathrm{Common}\mathrm{angle}\right)$

$\frac{PQ}{PA}=\frac{PR}{PB}=\frac{2}{1}$

So, by SAS similarity, we get

$△PQR~△PAB$

Thus, the corresponding angles are equal.

$\therefore \angle PQR=\angle PAB$ and $\angle PRQ=\angle PBA$

$\Rightarrow AB\parallel QR$

Hence, proved.

Step 3: Prove that the line segment joining the mid-points of two sides of a triangle is one-half of the third side:

$△PAB~△PQR$

Thus, the corresponding sides are proportional.

$\frac{PA}{PQ}=\frac{AB}{QR}=\frac{PB}{PR}$

$\Rightarrow \frac{AB}{QR}=\frac{PA}{PQ}$

$\Rightarrow \frac{AB}{QR}=\frac{PA}{PA+AQ}$

$\Rightarrow \frac{AB}{QR}=\frac{PA}{PA+PA}$ …… $\left(A\text{is mid-point of}PQ\right)$

$\Rightarrow \frac{AB}{QR}=\frac{PA}{2PA}$

$\Rightarrow \frac{AB}{QR}=\frac{1}{2}$

$\Rightarrow AB=\frac{1}{2}QR$

Hence, proved.