Question

In ΔABC, AM is a median and N is the mid-point of AM. BN produced meets AC at P. Prove that AP = `1/3` AC.

[Hint: Through M draw MQ parallel to BP.]

Answer 1

Step 1:

Draw MQ parallel to BP (or NP) such that Q lies on AC

Step 2:

In ΔAMQ, N is the midpoint of AM

NP || MQ (since BNP || MQ)

By the converse of the midpoint theorem, P is the midpoint of AQ

So, AP = PQ

Step 3:

In ΔBPC, M is the midpoint of BC (since AM is a median)

MQ || BP

By the converse of the midpoint theorem, Q is the midpoint of PC

So, PQ = QC

Step 4:

From Step 2, AP = PQ

From Step 3, PQ = QC

Therefore, AP = PQ = QC

AC = AP + PQ + QC

Substitute AP for PQ and QC: AC = AP + AP + AP = 3AP

Thus, `AP = 1/3 AC`.