M and N divide the side of AB of triangle ABC into 3 equal parts. Line segments MP and NQ are both parallel to BC and meets at AC at P and Q respectively. Prove P and Q divide AC into 3 equal parts.

Hi..!!!

Here is the answer to your question.

**Given :** A triangle ABC in which M and N divides AB into three equal parts i.e. AM = MN = NB and MP || NQ || BC.

**To prove:** P and Q divide AC into three equal parts i.e. AP = PQ = QC

**Construction **Join MC. Let it intersects QN at R.

**Proof**: We have AM = MN = NB

⇒ AM = MN

⇒ M is the mid point of AN.

Consider Δ ANQ,

M is the mid point of AN and MP || NQ (Given)

∴ By mid point theorem, P is the mid point of AQ

⇒ AP = PQ .....(1)

Consider ΔBCM, N is the mid point of MB (as MN = NB) and NR || BC (as NQ || BC)

⇒ by mid point theorem , R is the mid point of MC.

Now, Consider Δ CPM,

R is mid point of MC ( proved above) and QR || PM (as QN || PM)

⇒ Q is mid point of CP (by mid point theorem)

⇒ CQ = QP

From (1) and (2)

AP = PQ = QC

⇒ P and Q divides AC into 3 equal parts.

Hence, the result is proved.

Best wishes..!!!