S and T are points on sides PR and QR of \(\triangle PQR\) such that \(\angle P = \angle RTS\). Show that \(\triangle RPQ \sim \triangle RTS\).
Given: In \(\triangle PQR\), point \(S\) lies on side \(PR\) and point \(T\) lies on side \(QR\). Also, \(\angle P = \angle RTS\).
To prove: \(\triangle RPQ \sim \triangle RTS\).
Since \(S\) lies on \(PR\), segment \(RS\) is a part of line \(PR\). Similarly, \(T\) lies on \(QR\), so segment \(RT\) is a part of line \(QR\).
Therefore, the angle between \(RS\) and \(RT\) is the same as the angle between \(PR\) and \(QR\):
\[\angle SRT = \angle PRQ.\]
But \(\angle PRQ\) is angle \(R\) of \(\triangle RPQ\). Hence,
\[\angle R \text{ (in } \triangle RPQ) = \angle SRT \text{ (in } \triangle RTS).\]
We are given that
\[\angle P = \angle RTS.\]
So angle at \(P\) in \(\triangle RPQ\) equals angle at \(T\) in \(\triangle RTS\).
Thus we have two pairs of equal angles:
\[\angle P = \angle RTS, \quad \angle R = \angle SRT.\]
Therefore, by the AA similarity criterion,
\[\triangle RPQ \sim \triangle RTS.\]
Hence proved.