If P,Q,R divide the side BC,CA and AB of ΔABC in the same ratio, prove that the centroid of the triangle ABC and PQR coincide.

Let P, Q, R be the midpoints of sides BC, CA and AB respectively

Construct a ΔPQR by joining these three midpoints of the sides.

This is called the medial triangle

Since, PQ, QR and PR are midsegments of BC, AB and AC respectively

So,

Since the corresponding sides are proportional

∴ ΔPQR ≅ ΔABC

Now, we have to prove that the centroid of the triangle ABC and PQR coincide.

For that we must show that the medians of ΔABC pass through the midpoints of three sides of the medial triangle ΔPQR.

Since PQ is a midsegment of ΔABC,

⇒ PQ || BC, so PQ || BR.

And since QR is a midsegment of AB,

⇒ AB || QR, so QR || PB.

By definition, a quadrilateral PQRB is a parallelogram.

The medians BQ and CP are in fact the diagonals of the parallelogram PQRB.

And we know that the diagonals of a parallelogram bisect each other, so PD = DR.

In other words, D is the midpoint of PR.

In the similar manner, we can show that F and E are midpoints of RQ and PQ respectively.

Hence, the centroid of the triangle ABC and PQR coincide.