In the figure given alongside, \( T \) is a point on the side \( P S \) produced of a parallelogram \( P Q R S \) and \( Q T \) intersects \( S R \) at \( O \). Show that \( \triangle P Q T \sim \triangle R O Q \). \( S \) is a point on the side \( O R \)

We wish to prove that \[ \triangle P Q T \sim \triangle R O Q. \] In the diagram we have a parallelogram \(P Q R S\) with the usual properties \[ P Q \parallel S R\quad\text{and}\quad P S \parallel Q R. \] Also, the point \(T\) lies on the line through \(P\) and \(S\) (with \(S\) between \(P\) and \(T\)) and the line \(Q T\) (extended) meets \(S R\) at \(O\). A key idea is to show that the angles of the two triangles coincide when we choose the correspondence \[ P\longleftrightarrow R,\quad Q\longleftrightarrow O,\quad T\longleftrightarrow Q. \] Once this correspondence is established, the proof is a matter of a few angle‐chasing steps. --- **Step 1. Show that the sides corresponding are parallel.** 1.1. In \(\triangle P Q T\) the side \(P Q\) corresponds to the side \(R O\) in \(\triangle R O Q\). Since \(P Q\) is a side of the parallelogram, and \(S R\) is opposite to it, we have \[ P Q \parallel S R. \] But \(O\) lies on \(S R\) (indeed, \(Q T\) meets \(S R\) at \(O\)). Hence, \(R O\) is part of \(S R\) and \[ P Q \parallel R O. \] 1.2. In \(\triangle P Q T\) the side \(P T\) corresponds to the side \(R Q\) in \(\triangle R O Q\). Notice that \(T\) lies on the extension of \(P S\) and so \[ P T\ \text{is collinear with }P S. \] But in a parallelogram, we have \[ P S \parallel Q R. \] Thus \[ P T \parallel Q R. \] Since \(R Q\) is just \(Q R\) (reversing the order does not affect the line), we deduce that \[ P T \parallel R Q. \] --- **Step 2. Compare the angles.** Under our correspondence the vertices match as follows: