1. **Problem statement:** Given a circle with center $O$ and points $Q$, $R$, and $P$ on the circumference, where $OR = OP$ and $OQ = PQ$, and the central angle $\angle QOR = 70^\circ$, find the angle $\angle OQR$. 2. **Key properties and formulas:** - Since $OR = OP$, triangle $ORP$ is isosceles with $\angle OPR = \angle ORP$. - Since $OQ = PQ$, triangle $OQP$ is isosceles with $\angle OQP = \angle OPQ$. - The central angle $\angle QOR = 70^\circ$ subtends the arc $QR$. - The inscribed angle $\angle OQR$ subtends the same arc $OR$. 3. **Step-by-step solution:** - The central angle $\angle QOR = 70^\circ$ subtends arc $QR$. - The inscribed angle $\angle OQR$ subtending the same arc $OR$ is half the measure of the central angle subtending that arc. - Since $OR = OP$, triangle $ORP$ is isosceles, so $\angle ORP = \angle OPR$. - The total angle around point $O$ is $360^\circ$, so the arc $QR$ is $70^\circ$, and the remaining arc $RPQ$ is $290^\circ$. - The inscribed angle $\angle OQR$ subtends arc $OR$, which is part of the $290^\circ$ arc. 4. **Calculate $\angle OQR$:** - The inscribed angle is half the measure of the arc it subtends. - Since $\angle QOR = 70^\circ$, the arc $QR$ is $70^\circ$. - The arc $OR$ is the remaining part of the circle, $360^\circ - 70^\circ = 290^\circ$. - Therefore, $\angle OQR = \frac{1}{2} \times 70^\circ = 35^\circ$. **Final answer:** $$\boxed{35^\circ}$$