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In a triangle $\text{ABC, AB= 6, BC= 8}$ and $\text{AC=10}.$ A perpendicular dropped from $\text{B}$, meets the side $\text{AC}$ at $\text{D}.$ A circles of radius $\text{BD}$ (with centre $\text{B})$ is drawn. If the circle cuts $\text{AB}$ and $\text{BC}$ at $\text{P}$ and $\text{Q}$ respectively, then $\text{AP : QC}$ is equal to

  1. $1:1$
  2. $3:2$
  3. $4:1$
  4. $3:8$
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