Idk if this is helpful or not but consider the circumcircle of ABC then let CP intersects circle at T then use cyclic I will edit it when I comeup with my solution

Let $S$ be the point such that $ABSC$ is a parallelogram. Then $AP + CS = PC$ (call this $(*)$) and $\tfrac{AP}{CS} = \tfrac{PR}{RC}$, so $AP = PR$ and $CR = CS$. Subtracting from $(*)$ gives $CR = AB$.

Let $S$ be the point such that $ABSC$ is a parallelogram. Then $AP + CS = PC$ (call this $(*)$) and $\tfrac{AP}{CS} = \tfrac{PR}{RC}$, so $AP = PR$ and $CR = CS$. Subtracting from $(*)$ gives $CR = AB$.

Let $B'$ be the reflection of $B$ over $A$. Then $\triangle PB'C$ is isosceles and $AQ\parallel B'C$, so $\triangle PAR$ is isosceles and we have $$CR = CP - RP = B'P - AP = B'A = AB$$ as desired.

notice ARQ collinear then menelaus $\frac{PR}{RC} = \frac{AP}{BA}$ $\frac{AP}{AB} = \frac{PR}{RC} = k$ By First Condition $(k+1)AB = (k+1) RC$ $AB=RC$