In $$\triangle$$ABC, the line parallel to $$BC$$ intersects $$AB$$ and $$AC$$ at $$P$$ and $$Q$$ respectively. If $$AB:AP=5:3$$, then $$AQ:QC$$ is:

Given : PQ $$\parallel$$ BC and $$AB:AP=5:3$$

To find : $$AQ:QC$$

Solution : In $$\triangle$$ APQ and $$\triangle$$ ABC,

$$\angle$$ APQ = $$\angle$$ ABC   (Corresponding angles)

$$\angle$$ AQP = $$\angle$$ ACB   (Corresponding angles)

=> $$\triangle$$ APQ $$\sim$$ $$\triangle$$ ABC (By AA criterion)

=> $$\frac{AQ}{QC}=\frac{AP}{PB}$$

=> $$\frac{AQ}{QC}=\frac{AP}{AB-AP}$$

$$\frac{AQ}{QC}=\frac{3}{2}$$

=> Ans - (A)