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.
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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.