POINTS M AND N DIVIDE THE SIDE AB OF TRAINGLE ABC INTO THREE EQUAL PARTS. LINE SEGMENTS MP AND NQ ARE BOTH PARALLEL TO BC AND MEET AC IN P AND Q RESPECTIVELY. PROVE THAT P AND Q DIVIDE AC INTO THREE EQUAL PARTS.

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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.