1. **Problem statement:** In triangle $\triangle PQR$, point $S$ lies on side $QR$ such that $PS = SR = SQ$. Given that $\angle PRS = 55^\circ$, find $\angle SQP$. 2. **Understanding the problem:** Since $PS = SR = SQ$, point $S$ is equidistant from $P$, $Q$, and $R$ in a special way. This implies $S$ is the center of an equilateral configuration on $QR$ and $PS$ is equal to the segments $SR$ and $SQ$. 3. **Key observations:** - $S$ lies on $QR$. - $SQ = SR$ means $S$ is the midpoint of $QR$. - $PS = SR = SQ$ means triangle $PSQ$ and $PSR$ are isosceles with two equal sides. 4. **Given:** $\angle PRS = 55^\circ$. 5. **Goal:** Find $\angle SQP$. 6. **Step-by-step solution:** - Since $S$ is midpoint of $QR$, $SQ = SR$. - Given $PS = SR = SQ$, triangles $PSQ$ and $PSR$ are isosceles with $PS = SQ$ and $PS = SR$ respectively. - In $\triangle PRS$, $PS = SR$, so $\triangle PRS$ is isosceles with $PS = SR$. - Given $\angle PRS = 55^\circ$, the base angles $\angle RPS$ and $\angle RSP$ are equal. - Sum of angles in $\triangle PRS$ is $180^\circ$: $$\angle PRS + 2 \times \angle RPS = 180^\circ$$ $$55^\circ + 2 \times \angle RPS = 180^\circ$$ $$2 \times \angle RPS = 125^\circ$$ $$\angle RPS = 62.5^\circ$$ - Since $PS = SQ$, $\triangle PSQ$ is isosceles with $PS = SQ$. - We want to find $\angle SQP$, which is the angle at $Q$ in $\triangle PSQ$. - In $\triangle PSQ$, sum of angles is $180^\circ$: $$\angle SQP + \angle QPS + \angle PQS = 180^\circ$$ - Since $PS = SQ$, angles opposite these sides are equal: $$\angle SQP = \angle PQS$$ - Let $\angle SQP = \angle PQS = x$. - Then: $$x + x + \angle QPS = 180^\circ$$ $$2x + \angle QPS = 180^\circ$$ - Note that $\angle QPS$ is the same as $\angle RPS$ because points $Q$, $S$, and $R$ are collinear and $S$ is midpoint. - So, $\angle QPS = 62.5^\circ$. - Substitute: $$2x + 62.5^\circ = 180^\circ$$ $$2x = 117.5^\circ$$ $$x = 58.75^\circ$$ 7. **Final answer:** $$\boxed{\angle SQP = 58.75^\circ}$$