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Consider a triangle $ABC$.The sides AB and AC are extended to points D and E, respectively, such that $AD = 3AB$ and $AE = 3AC$.Then one diagonal of $BDEC$ divides the other diagonal in the ratio?

I got the answer 1:3 but assuming the triangle isosceles and taking A(0,0), B(-1,-1) and C(1,-1). But how to approach the question without assuming anything?

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Let $O$ be the point of intersection of diagonals. Consider $\triangle ABC$ and $\triangle ADE$ they are similar (why?). So if you consider $BCO$ and $DEO$ they are similar too, so $OE/OB = DE/BC = 3$.

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