Question

Medians CM and BN of ΔABC are produced to P and Q respectively such that CM = MP and BN = NQ. Prove that the points P, A, Q are collinear and PA = AQ.

[Hint: Join MN.MN = `1/2` PA and || to PA in ΔCAP, MN = `1/2` AQ and || to AQ in ΔBAQ, etc]

Answer 1

Given:

• ΔABC is a triangle with medians CM and BN,
• M is the midpoint of AB,
• N is the midpoint of AC,
• The medians CM and BN are extended to points P and Q, respectively, such that:
  • CM = MP,
  • BN = NQ.

We need to prove:

1. The points P, A and Q are collinear.
2. PA = AQ.

Step 1: Use of the centroid property

In a triangle, the medians meet at the centroid G, which divides each median into two parts in the ratio 2 : 1, with the longer segment being closer to the vertex.

Thus, the centroid G divides:

• CM such that CG = `2/3` × CM and GM = `1/3` × CM,
• BN such that BG = `2/3` × BN and GN = `1/3` × BN.

Step 2: Extend the medians to points P and Q

We are given that CM = MP and BN = NQ, meaning the points P and Q are such that:

• P is on the line through C and M and the length of segment MP is equal to the length of segment CM,
• Q is on the line through B and N and the length of segment NQ is equal to the length of segment BN.

Step 3: Investigating the relationship of the points P, A and Q

Since CM = MP and BN = NQ, the points P and Q are reflections of M and N across the centroid G. Therefore, the centroid G lies on the line joining P and Q and it divides this line segment in a 1 : 1 ratio.

Thus, points P, A and Q are collinear because G divides PQ symmetrically and lies along the line joining these points.

Step 4: Prove that PA = AQ

Since G is the centroid of triangle ABC and it divides the line segment joining P and Q symmetrically, we have:

• PA = AQ, as the centroid divides the segment joining the points P and Q into two equal lengths.