Solution :
Given : \(\triangle\) RPQ and \(\triangle\) RTS where \(\angle\) P = \(\angle\) RTS
To prove : \(\triangle\) RPQ ~ \(\triangle\) RTS
Proof : In \(\triangle\) RPQ and \(\triangle\) RTS
Given, \(\angle\) P = \(\angle\) RTS
\(\angle\) R = \(\angle\) R (common)
Hence, By AA similarity, \(\triangle\) RPQ ~ \(\triangle\) RTS