In the triangle ΔABC, the two medians BP and CQ are so extended upto the points R and S respectively such that BP = PR and CQ = QS. Let us prove that, S, A, R are collinear.

The figure according to given information is as shown below

Consider ΔAQS and ΔBQC

QS = QC … given

∠SQA = ∠CQB … vertically opposite angles

AQ = BQ … CQ is median on AB

Therefore, ΔAQS ≅ ΔBQC … SAS test for congruency

⇒ ∠ASQ = ∠BCQ … corresponding angles of congruent triangles

Thus AS || BC because ∠ASQ and ∠BCQ are pair of alternate interior angles with transversal as CS

⇒ AS || BC … (i)

Consider ΔAR and ΔCPB

BP = PR … given

∠APR = ∠BPC … vertically opposite angles

AP = CP … BP is median on AC

Therefore, ΔAPR ≅ ΔCPB … SAS test for congruency

⇒ ∠ARP = ∠CBP … corresponding angles of congruent triangles

Thus AR || BC because ∠ARP and ∠CBP are pair of alternate interior angles with transversal as BR

⇒ AR || BC … (ii)

From (i) and (ii) we can say that

AS || AR

But point A lies on both the lines AS and AR which means AS and AR are on the same straight line

Thus, point A, S and R are collinear points