Question : $\triangle$ABC is similar to $\triangle$PQR and AB : PQ = 2 : 3. AD is the median to the side BC in $\triangle$ABC and PS is the median to the side QR in $\triangle$PQR. What is the value of $(\frac{BD}{QS})^2$?
Option 1: $\frac{3}{5}$
Option 2: $\frac{4}{9}$
Option 3: $\frac{2}{3}$
Option 4: $\frac{4}{7}$
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Correct Answer: $\frac{4}{9}$
Solution : AB : PQ = 2 : 3 As $\triangle$ABC is similar to $\triangle$PQR, $\frac{AB}{PQ}$ = $\frac{BC}{QR}$ = $\frac{CA}{RP}$ = $\frac{AD}{PS}$ = $\frac{BD}{QS}$ = $\frac{DC}{SR}$ = $\frac{2}{3}$ $(\frac{BD}{QS})^2$ = $(\frac{2}{3})^2$ = $\frac{4}{9}$ Hence, the correct answer is $\frac{4}{9}$.
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