In a ΔABC, P and Q are points on AB and AC respectively, such that AP = 3 cm, PB = 6 cm, AQ = 5 cm and QC = 10 cm, then BC = 3 PQ.TrueFalse

The given statement is True.

Here's the step-by-step solution:

Given that in triangle ABC, P and Q are points on AB and AC respectively.

AP = 3 cm, PB = 6 cm, AQ = 5 cm, QC = 10 cm.

According to the section formula, if a point divides a line segment internally in the ratio m:n, then the point is (mx2 + nx1)/(m+n), (my2 + ny1)/(m+n).

Here, point P divides AB in the ratio 1:2 (since AP = 3 cm and PB = 6 cm) and point Q divides AC in the ratio 1:2 (since AQ = 5 cm and QC = 10 cm).

Therefore, by the converse of the Basic Proportionality Theorem (also known as Thales Theorem), the line PQ is parallel to BC.

And, by the Basic Proportionality Theorem, if a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides those sides in the same ratio.

So, BC = 3 PQ. Hence, the given statement is true.