In the given triangle P, Q and R are mid-points of sides AB, BC and AC respectively. Prove that triangle PQR is similar to triangle ABC.

In ∆ABC,

Since, P is mid-point of AB and R is mid-point of AC so,

By mid-point theorem,

PR || BC.

By basic proportionality theorem we have,

$\dfrac{AP}{PB} = \dfrac{AR}{RC}$

And, in ∆PAR and ∆BAC,

∠PAR = ∠BAC [Common]

∠APR = ∠ABC [Corresponding angles are equal]

∴ ∆PAR ~ ∆BAC [By AA]

Since, corresponding sides of similar triangles are proportional.

$\Rightarrow \dfrac{PR}{BC} = \dfrac{AP}{AB}$

$\Rightarrow \dfrac{PR}{BC} = \dfrac{1}{2}$ [∵ P is the mid-point of AB]

$\Rightarrow PR = \dfrac{1}{2}BC$

Since, P and Q are mid-points of AB and BC so, by mid-point theorem,

PQ = $\dfrac{1}{2}$AC

Since, R and Q are mid-points of AC and BC so, by mid-point theorem,

RQ = $\dfrac{1}{2}$AB

So,

$\dfrac{PR}{BC} = \dfrac{PQ}{AC} = \dfrac{RQ}{AB} = \dfrac{1}{2}$

Hence, proved that ∆PQR ~ ∆ABC by SSS similarity.