Respuesta :

The diagram can be drawn as,

In triangle PQR and PST

[tex]\begin{gathered} \angle QPR=\angle TPS(V.O.A) \\ \frac{QP}{PS}=\frac{9}{3}=3 \\ \frac{RP}{PT}^{}=\frac{6}{2}=3 \\ \frac{QP}{PS}=\frac{RP}{PT} \end{gathered}[/tex]

Thus, from SAS,

[tex]\Delta STP\approx\Delta QRP[/tex]

Thus, option (A) is the correct solution.

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