[tex]In\text{ }\Delta PQS\text{ and }\Delta RQS[/tex]
[tex]\bar{PQ}\cong\bar{RQ}[/tex][tex]\angle PQS=\angle RQS\text{ (since QS bisects }\angle PQR)[/tex][tex]\bar{QS}=\bar{QS}\text{ (common)}[/tex]
Therefore,
[tex]\Delta PQS\cong\Delta RQS\text{ (SAS congruence rule.)}[/tex]
So,
[tex]\angle P\cong\angle R[/tex]
