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In two $\triangle$s ABC and PQR, if $\frac{AB}{QR} = \frac{BC}{QP} = \frac{AC}{PR}$, then
- (a)$\triangle PQR \sim \triangle CAB$
- (b)$\triangle PQR \sim \triangle ABC$
- (c)$\triangle PQR \sim \triangle CBA$
- (d)$\triangle PQR \sim \triangle BCA$
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Sol. (C) $\triangle PQR \sim \triangle CBA$