$ \triangle PQR \ $ is an isosceles triangle where $PQ = PR. $ $X$ is a point on the circumcircle of $ΔPQR $, such that it lies on the arc $QR$. The normal drawn from the point $P$ on $XR$ intersects $XR$ at point $Y.$ Prove that $$QX + YR = XY.$$
Reflect the $R$ across line $PY$ in to $Z$. So $Z\in XR$ and $YZ = YR$ and $$PZ = PR = PQ$$
Sp we have to prove that $QX = XZ$. Since $$\angle XZP = \pi-\angle RZP = \pi-\angle XRP = \angle XQP$$ and $$\angle QXP = \angle QRP = \angle RQP = \angle ZXP$$ we see that triangle $XQP$ is congruent to triangle $XZP$ (s.a.s.) and we are done.