Question

in a triangle ABC ,P and Q are the midpoints of side AB and AC respectively as shown in the figure R and S are the midpoints of PCand PB respectively prove that BQ and SR bisect each other

Answer 1

Following Basic Proportionality theorem, as AP/PB = AQ/QC = 1, PQ || BC
Again PR/RB = RS/SC = 1 ensures RS || BC
Now in triangle BQP, as OR || PQ, by converse of BPT, BR/RP = BO/OQ = 1
==> BO = OQ
Thereby O is the mid point of BQ
So RS bisects BQ [PROVED]
NOTE: "each other" seems impossible

Answer 2

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