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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
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B
3:4
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C
5:4
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D
4:5
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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 $$2:1$$

$$\implies BG=\dfrac{2}{3}BS=\dfrac{2}{3}\left(\dfrac{2}{3}x\right)=\dfrac{4}{9}x$$

$$GS=\dfrac{1}{3}BS=\dfrac{1}{3}\left(\dfrac{2}{3}x\right)=\dfrac{2}{9}x$$

$$GR=GS+SR=\dfrac{1}{3}x+\dfrac{2}{9}x=\dfrac{5}{9}x$$

$$\implies \dfrac{BG}{RG}=\dfrac{\dfrac{4}{9}x}{\dfrac{5}{9}x}=\dfrac{4}{5}$$

$$\implies BG:RG=4:5$$

933808_395397_ans_54c73de16f5f484796fafd4ee96dcfb3.JPG

Mathematics

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