[tex]\angle QPR\text{ = }\angle SRT\text{ (Corresponding of F angles)}[/tex][tex]\begin{gathered} \angle PQR\text{ +}\angle QPR+\angle QRP=180^o=\angle RST+\angle STR+\angle SRT\text{ (Sum of angles in a triangle)} \\ \Rightarrow\angle PQR\text{ +}\angle QPR+\angle QRP=\angle RST+\angle QRP+\angle QPR \\ \Rightarrow\angle PQR=\angle RST \\ \text{therefore triangle PQR is similar to triangle RST} \\ \text{But magnitude of line QP is equal to magnitude of line }SR \\ \text{Therefore} \\ triangle\text{ PQR }\cong\text{ triangle RST} \end{gathered}[/tex]
