如图所示，圆 BDEFG 的圆心为 O，直径为 BE。直线 AC 切圆于点 B，且 $\angle BOD = 30^\circ$。角 EBF、角 FBG 与角 GBC 的比为 1 : 1 : 2。

1. Relate the equal angles at $B$ to their intercepted arcs

Given
$\angle EBF : \angle FBG : \angle GBC = 1:1:2,$
so $\angle EBF = \angle FBG$.

Equal angles at the circumference stand on equal arcs, hence arc $EF =$ arc $FG$.

1. Relate the angle at the centre to the angle at the circumference standing on chord $DB$

$\angle BOD = 30^\circ$.

Since $\angle DEB$ and $\angle BOD$ stand on the same arc $DB$,
$\angle DEB = \frac{1}{2}\angle BOD = \frac{1}{2}(30^\circ)=15^\circ.$

1. Find $\angle EBG$ from the given ratio

Since $AC$ is tangent at $B$ and $BE$ is a diameter, $BE$ is along the radius $OB$.

So $\angle EBC = 90^\circ$.

Given
$\angle EBF : \angle FBG : \angle GBC = 1:1:2,$
let the angles be $x$, $x$ and $2x$.

$x+x+2x=90^\circ$
$4x=90^\circ$
$x=22.5^\circ$

Hence
$\angle EBG=\angle EBF+\angle FBG=22.5^\circ+22.5^\circ=45^\circ.$

2. Use cyclic quadrilateral property to find $\angle EFG$

Points $E$, $F$, $G$ and $B$ lie on the same circle, so $EBGF$ is a cyclic quadrilateral.

Opposite angles of a cyclic quadrilateral are supplementary:
$\angle EFG + \angle EBG = 180^\circ.$

$\angle EFG = 180^\circ - 45^\circ = 135^\circ.$