S and T are points on sides PR and QR of PQR such that P = RTS. Show that RPQ ~ RTS.
Step-by-Step Solution
Key Concept: Use the AA (Angle‑Angle) similarity criterion. Because S lies on \(PR\) and T lies on \(QR\), the lines \(RS\) and \(RT\) are respectively collinear with \(PR\) and \(QR\). Hence \(\angle PRQ = \angle R\) is common to both triangles. Together with the given equality \(\angle P = \angle RTS\), we have two equal angles, which is sufficient to conclude similarity.
1. Identify the given information
- \(S\) lies on side \(PR\) and \(T\) lies on side \(QR\) of \(\Delta PQR\).
- \(\angle P = \angle RTS\) is given.
2. Observe the collinearity
- Since \(S\) is on \(PR\), the line \(RS\) is the same as line \(RP\).
- Since \(T\) is on \(QR\), the line \(RT\) is the same as line \(RQ\).
3. Find a common angle
- In \(\Delta RPQ\) the angle at vertex \(R\) is \(\angle PRQ\).
- In \(\Delta RTS\) the angle at vertex \(R\) is \(\angle SRT\) (formed by \(RS\) and \(RT\)).
- Because \(RS\) lies on \(RP\) and \(RT\) lies on \(RQ\), we have \(\angle PRQ = \angle SRT\). Thus \(\angle R\) is common to both triangles.
4. Apply the AA similarity criterion
- We have \(\angle P = \angle RTS\) (given).
- We have \(\angle R\) common to both triangles.
- Therefore, by AA, \(\Delta RPQ \sim \Delta RTS\).
5. State the corresponding sides (optional for full marks)
- Correspondence: \(R \leftrightarrow R\), \(P \leftrightarrow T\), \(Q \leftrightarrow S\).
- Hence \(\frac{RP}{RT} = \frac{RQ}{RS} = \frac{PQ}{TS}\).
Correct Answer: Since \(\angle P = \angle RTS\) and \(\angle PRQ = \angle SRT\) (common angle), by AA criterion \(\Delta RPQ \sim \Delta RTS\).