Question

# $$P, Q$$ and $$R$$ are on $$AB, BC$$ and $$AC$$ of the equilateral triangle $$ABC$$ respectively. $$AP: PB = CQ: QB = 1: 2$$. $$G$$ is the centroid of the triangle $$PQB$$ and $$R$$ is the midpoint of $$AC$$. Find $$BG: GR$$.

A
4:3
B
3:4
C
5:4
D
4:5

Solution

## The correct option is D $$4: 5$$Given,$$BQ:CQ=2:1$$ and $$BP:PB=2:1$$Therefore by converse of Basic proportionality theorem,$$PQ\parallel AC$$$$\implies SQ\parallel RC$$$$\implies BS:SR=2:1$$   [Basic Proportionality Theorem]Let $$BR=x$$$$BS=\dfrac{2}{3}x$$ and$$SR=\dfrac{1}{3}x$$$$G$$ is the centroid of the $$\triangle PQB$$ and centroid divides median in $$2:1$$$$\implies BG=\dfrac{2}{3}BS=\dfrac{2}{3}\left(\dfrac{2}{3}x\right)=\dfrac{4}{9}x$$$$GS=\dfrac{1}{3}BS=\dfrac{1}{3}\left(\dfrac{2}{3}x\right)=\dfrac{2}{9}x$$$$GR=GS+SR=\dfrac{1}{3}x+\dfrac{2}{9}x=\dfrac{5}{9}x$$$$\implies \dfrac{BG}{RG}=\dfrac{\dfrac{4}{9}x}{\dfrac{5}{9}x}=\dfrac{4}{5}$$$$\implies BG:RG=4:5$$Mathematics

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