In Fig. 6.16, $\dfrac{PS}{SQ} = \dfrac{PT}{TR}$ and $\angle PST = \angle PRQ$. Prove that PQR is an isosceles triangle. [Figure: triangle PQR with vertex P at top; points S on PQ and T on PR with segment ST drawn; angle PST is marked at S.]

Step 1: State the given information

We are given two conditions. The first condition describes the ratio of the segments on sides PQ and PR, and the second condition states that angle PST is equal to angle PRQ.

Step 2: Apply the converse of Basic Proportionality Theorem

According to the converse of the Basic Proportionality Theorem (also known as Thales Theorem or Thales's Theorem), if a line divides two sides of a triangle proportionally, then the line is parallel to the third side. Here, line segment ST divides sides PQ and PR of triangle PQR proportionally, so ST must be parallel to QR.

Step 3: Identify corresponding angles

When two parallel lines (ST and QR) are intersected by a transversal line (PQ), the corresponding angles formed are equal. Therefore, angle PST is equal to angle PQR.

Step 4: Use the given angle equality

We are given that $\angle PST = \angle PRQ$. From the previous step, we established that $\angle PST = \angle PQR$. By substituting $\angle PST$ with $\angle PQR$ in the given condition, we conclude that $\angle PQR = \angle PRQ$.

Step 5: Conclude that triangle PQR is isosceles

In any triangle, if two angles are equal, then the sides opposite to these equal angles are also equal. Since $\angle PQR = \angle PRQ$ in $\triangle PQR$, the side opposite to $\angle PQR$ (which is PR) must be equal to the side opposite to $\angle PRQ$ (which is PQ). This means $\triangle PQR$ is an isosceles triangle with PQ = PR.

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