In triangle ABC ,the medians BP and CQ are produced upto points M and N respectively such that BP =PM and CQ =QN .prove that :
1) M , N and A are collinear.
2) A is the mid point of MN.

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Answer:

Given: P and Q are mid points of the sides AC and AB respectively.Extend BP to M and CQ to N such that, BP = PM and CQ = QNTo prove:N, A and M are collinear and A is mid point of MNProof: ∆APM and △BPCPM = BPAP = PC∠APM=∠BPC (Vertically oppposite angle)△APM≅△BPC (By SAS)∠AMP =∠PBC (by CPCT)AM = BC (by CPCT)Which is a pair of alternate interior angle so, AM ∥BC−−−−(1)Similarly, △ANQ≅△BQC (By SAS)∠ANQ =∠QBC (by CPCT)AN = BC (by cpct)Which is a pair of alternate interior angle so, AN∥BC−−−−(2)From (1) and (2)AM is a straight line so N,A and M are collinearAnd NM = NA+AM = BC +BC = 2BC So, A is the mid point of MNHence proved

 
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