Problem 1455

1. Identify Key Points and Properties:
- Given triangle $ABC$ with $CA = CB$.
- Point $E$ lies on the circumcircle of $ABC$ such that $\angle ECB = 90^\circ$.
- Line through $E$ parallel to $CB$ intersects $CA$ at $F$ and $AB$ at $G$.

2. **Establish Rectangle $BCEP$:**
- Let $P$ be the other intersection of $EF$ with the circumcircle of $ABC$.
- Since $\angle ECB = 90^\circ$, quadrilateral $BCEP$ is a rectangle.
- Therefore, $EP = BC$ and $CP = BE$.

3. Power of a Point and Segment Lengths:
- By the Power of a Point theorem, $FE$ equals one of $AF$ or $FC$.
- Since $\angle FEC = 90^\circ$, $FC > FE$, so $FE = AF$ and $PF = FC$.

4. Angle Relationships and Arc Midpoint:
- $\angle BCP = \angle EBC = \angle EAC = \angle AEP = \angle ACP$.
- Thus, $P$ is the arc midpoint of minor arc $AB$, and $P$ lies on the perpendicular bisector of $AB$.
- Since $C$ lies on the perpendicular bisector of $AB$, $CP \perp AB$.

5. Isosceles Trapezoid and Cyclic Quadrilateral:
- Let $Q$ be the point on line $BC$ such that $BE \perp QE$.
- $\angle EQB = 90^\circ - \angle EBC = 90^\circ - \angle PCB = \angle GBQ$.
- Since $BQ \parallel EG$, quadrilateral $BQEG$ is an isosceles trapezoid and thus cyclic.

6. **Circumcenter $O$ of Triangle $BEG$:**
- Let $O$ be the circumcenter of triangle $BEG$.
- Since $\angle BEQ = 90^\circ$, $O$ is the midpoint of $BQ$ and $OB = OE$.

7. **Cyclic Quadrilateral $CEFO$:**
- Since $PF = FC$, $\angle EFC = 2\angle EPC = 2\angle EBC = \angle EOC$.
- Therefore, quadrilateral $CEFO$ is cyclic (in fact, it is a rectangle).

Thus, the center of the circumcircle of triangle $EGB$ lies on the circumcircle of triangle $ECF$.

$\blacksquare$