$\bigcirc Q, \bigcirc P$ is tangent with $\bigcirc O$ at point $A, B$. $\bigcirc Q, \bigcirc P$ intersect at $C,D$. Line $AD$ is intersects $\bigcirc O$ in $E$. Line $DB$ is intersects $\bigcirc O$ in $G$. Line $CA, CB$ is intersect $\bigcirc O$ in $S$ resp. $T$. Find if $\frac{EF}{SF}=\frac{GF}{TF}$ (with graph below)
The purpose of the two small tangential circles is to ensure that the three bold-marked lines are parallel. See figure #1.
If $DF$ cuts the circle $ESFTG$ at $Y$, then we have 3 isosceles trapeziums with $YG = FT$, $YE = FS$ and $EG = ST$.
Then, we have (1) $\triangle YEG \cong \triangle FST$; (2) $\stackrel{\frown}{YG} = \stackrel{\frown}{FT}$, and $\stackrel{\frown}{YE} = \stackrel{\frown}{FS}$. If we let $M$ and $N$ be the midpoints of $\stackrel{\frown}{ES}$ and $\stackrel{\frown}{GT}$ respectively, clearly we have ($\stackrel{\frown}{EM} = \stackrel{\frown}{MS}$; $\stackrel{\frown}{GN} = \stackrel{\frown}{NT}$). Note also that $MN$ is the diameter of the circle $ESFTG$ and it cuts $ES$, $YF$ and $GT$ at right angles. See figure #2.
By considering the sums of equal arcs again, $MG$ and $NE$ are the angle bisectors of $\angle FGY$ and $\angle FEY$ respectively.
Proving the equality of said ratios is the same as proving whether $\dfrac {EY}{EF} =\dfrac {GY}{GF}$ because of the proved congruence.
Let (1) CF cut MN at K; (2) MG intersect EN at O; (3) ME extended cut $FKYDC$ at $Z_1$; and (4) NG extended intersect $FKYDC$ at $Z_2$.
We digress to prove that (1) $Z_1 = Z_2 = Z$; and (2) ZK is an altitudes of $\triangle MNZ$.
“$\angle 1= \angle 2 = \angle3 = \angle 4$” implies $G, O, E, Z_1$ are con-cyclic. Similarly, $G, O, E, Z_2$ are also con-cyclic. This proves the above claim. See figure #3.
By the converse of the angle bisector theorem, we are done if MG, EN and ZK are concurrent. They will because all of them are altitudes of $\triangle ZMN$ and they are concurrent at O, the ortho-center.