In fig. 6. 36, QR/QS =QT/PR and.‹1=‹2.Show that triangle PQS is similar to triangle TQR

In fig. 6. 36, QR/QS =QT/PR and.‹1=‹2.Show that triangle PQS is si

1.	In fig. 6. 36, QR/QS =QT/PR and.‹1=‹2.Show that triangle PQS is similar to triangle TQR
Answer» Given:\xa0In figure, {tex}\\frac{{QR}}{{QS}} = \\frac{{QT}}{{PR}}{/tex}and {tex}\\angle{/tex} 1 = {tex}\\angle{/tex} 2To prove: {tex}\\triangle PQS \\sim \\triangle TQR{/tex}Proof: In {tex}\\triangle {/tex} PQR {tex}\\because {/tex}{tex}\\angle{/tex} 1= {tex}\\angle{/tex} 2{tex}\\therefore {/tex} PR = QP (1).......[ {tex}\\because {/tex} sides opposite to equal angle of a triangle are equal]Now, {tex}\\frac{{QR}}{{QS}} = \\frac{{QT}}{{PR}}{/tex} ......given{tex} \\Rightarrow \\frac{{QR}}{{QS}} = \\frac{{QT}}{{QP}}{/tex} (2).......Using(1)Again in {tex}\\triangle PQS{/tex} and {tex}\\triangle TQR{/tex}{tex}\\because \\frac{{QR}}{{QS}} = \\frac{{QT}}{{QP}}...........From(2){/tex}{tex}\\therefore \\frac{{QS}}{{QR}} = \\frac{{QP}}{{QT}}{/tex} and {tex}\\angle SQP = \\angle RQT{/tex}{tex}\\therefore \\triangle PQS \\sim \\triangle TQR{/tex}...........SAS similarity criterion