Prove PR = racPW imes QRQE using similar triangles.

Here are the proofs for the given statements:

11.2.1 Prove $PR = \frac{PW \cdot QR}{QE}$

Step 1: Identify similar triangles using the parallel lines.
Given that $PQ \parallel WE$.
Consider $\triangle PQR$ and $\triangle EWR$.
Since $PQ \parallel WE$, and $PRW$ is a transversal, $\angle QPR = \angle WPR$ (common angle). No, this is incorrect.
Since $PQ \parallel WE$, and $QW$ is a transversal, $\angle PQR = \angle EWR$ (alternate interior angles). No, this is incorrect.
Since $PQ \parallel WE$, and $PRW$ is a transversal, $\angle Q