Let $I$ be the second point of intersection of $\omega$ and $AC$,we obtain that $DI\parallel BC,BD=CI$,and hence \begin{align*}\frac{BH}{CH}& =\frac{\sin \angle BAH}{\sin \angle CAH}\\& =\frac{\sin \angle ABF\cdot BG}{AG}\cdot \frac{AG}{\sin \angle ACD\cdot CG}\\& =\frac{DF}{DI}\cdot \frac{CI}{CF}\\& =2\end{align*} as desired. $Q.E.D.$

Cross ratio bash: Let $M$ denote the midpoint of $BC$, $AM\cap CD=L$, It suffices to prove that $A(B,H;M,C)=-1$ $\iff -1=A(D,G;L,C)=B(D,G;L,C)$ which is true since $BL\cap EC\in \omega$ by symmetry and the harmonic division is well known.

To explain XmL's last step, let $ I = BL \cap AC $. Clearly $ I \in \omega $. To prove that $ (D, G; L, C) = -1 $, taking perspective at $ B $ and projecting onto $ \omega $ we find that it suffices to show that $ DIFC $ is a harmonic quadrilateral. So it suffices to show that $ DI * FC = DC * IF $. Now since $ \triangle{EIF} \sim \triangle{EFC} $ and $ \triangle{EID} \sim \triangle{EDC} $ we have that $ \frac{FI}{FC} = \frac{EF}{EC} = \frac{ED}{EC} = \frac{DI}{DC} $ as desired.

Let $D'$ be analogous to $D$ on $AC$, which is also on circle $BDC$. Then $DD'FC$ is a harmonic quadrilateral and so $AD'YC$ is a harmonic bundle where $Y=BF \cap AC$. Now let $AD=AD'=y,D'Y=y,YC=z$ and notice that $y/z=x/(x+y+z)$ by the harmonic bundle and so $xy+y^2+yz=xz$ and so $2xz=xz+zy+xy+y^2$ and so $\frac{z+y}{x} \times \frac{x+y}{z}=2$ and finish by Menelao.

let $AC \cap W=P$ . so we can find that $E(DFPC)=-1$ so we can find that $B(DFPC)=-1$ we know that $BF \cap DC=G$ we know that $PB \cap DC=N$ $B(DFPC)=-1$ so$(DGNC)=-1$ so$A(DGNC)=-1$ we know that $AM$ passes from $M$ the mid point of $BC$. so $(BHMC)=-1$ so we are done

Let $DG \cap BC =K$ and we tend to show $BC=KC$ let $I$ be the second intersection of $\omega$ .We have: $-1=(D,F;I,C)=(A,G;I,C)=(B,K;P_{\infty},C)$ Where the prespectives are taken WRT $B,D$ respectivly.And the claim clearly holds.Now since $(B,C;K,H)=-1$ we conclude $BH=2HC$. Edit:I made a typo in the solution and I am too lazy to fix it but you can take $G$ to be the intersection of $BF$ with $AC$ instead of the original defination to make the above solution make sense.

We will consider directed lengths. Let $AC\cap \omega=I$, $BI\cap AG=J$, and $IG\cap BC=K$. Clearly, $DI\|BC$ and $BD=IC$. Now, $$-1=(DF;IC)\stackrel{B}{=}(AG;JH)\stackrel{I}{=}(CK;BH).$$So, $\frac{CB}{CH}\div \frac{KB}{KH}=-1$. By Menelaus theorem on $\triangle ACH$ and $\triangle ABH$, we get \begin{align*} &\frac{AG}{GH}\cdot\frac{HK}{KC}\cdot\frac{CI}{IA}=-1~\text{and}~ \frac{AG}{GH}\cdot\frac{HC}{CB}\cdot\frac{BD}{DA}=-1 \\ \implies &\frac{HK}{KC}\cdot\frac{\cancel{CI}}{\cancel{IA}}\cdot\frac{CB}{HC}\cdot\frac{\cancel{DA}}{\cancel{BD}}=1\\ \implies &\frac{KH}{KC}\cdot\frac{CB}{CH}=1=-\frac{CB}{CH}\cdot\frac{KH}{KB}\\ \implies & KC=-KB\end{align*}Hence, $K$ is the midpoint of $BC$. Now, $$\frac{CB}{CH}=-\frac{KB}{KH}\implies \frac{2KC}{CH}=-\frac{KB}{KH}\implies HC=2KH.$$So, $BH=BK+KH=HC+2KH=2HC$. $\blacksquare$

Let $AC$ cut circle at $J$, $K$ is the midpoint of $BC$, $JB\cap CD=M$, $BF\cap CA=L$. $$-1=(D,F;J,C)=(D,G;M,C)=(P_{\infty},JG\cap BC;B,C)$$$\implies K$ is on $JG$. $$-1=(A,L;J,C)=(H,B;K,C)$$as desired.

Solution 1 - My Original Approach: Let $\omega$ meet $AC$ again at $D_1$, $M$ be the midpoint of $BC$, $D_1G$ meet $\omega$ again at $T$, and $A_1$ be the reflection of $A$ over $M$. Since $AB = AC$, we have $DD_1 \parallel BC$, so $AM, BD_1, CD$ concur at some point $K$. Trivially, $CDD_1F$ is a harmonic quadrilateral. Now, since inversion and reflection both preserve cross ratios, we know $$-1 = (C, D_1; D, F) \overset{G}{=} (D, T; C, B) \overset{D_1}{=} (D, G; C, K) \overset{A}{=} (AD \cap A_1C, AG \cap A_1C; C, A_1).$$Now, because $ABA_1C$ is a parallelogram, we know $AD \cap A_1C$ is at infinity, so $AG$ bisects $A_1C$. Hence, $H$ is the centroid of $ACA_1$, so $$\frac{BH}{HC} = \frac{BM + MH}{HC} = \frac{CM + \frac{CM}{3}}{\frac{2 \cdot CM}{3}} = \frac{\frac{4}{3}}{\frac{2}{3}} = 2$$as desired. $\blacksquare$ Solution 2 - More Slick: (We assume all properties and point labelings from the first paragraph of Solution $1$.) Let $X = BD_1 \cap AG$, $Y = BF \cap AC$, and $Z = XY \cap BC$. By Ceva-Menelaus wrt $ABC$, we have $$-1 = (H, DY \cap BC; B, C).$$On the other hand, notice $$-1 = (D, F; D_1, C) \overset{B}{=} (A, Y; D_1, C) \overset{X}{=} (H, Z; B, C)$$so $Z = DY \cap BC$, i.e. $D, X, Y, Z$ are all collinear. Now, observe $$-1 = (A, Y; D_1, C) \overset{D}{=} (B, Z; DD_1 \cap BC, C)$$so $C$ is the midpoint of $BZ$. But $-1 = (B, C; H, Z)$, so $$\frac{HB}{HC} = \frac{ZB}{ZC} = 2$$where the direction of lengths is irrelevant. $\blacksquare$ Remarks: Introducing $M$ is fairly natural due to the symmetry in this problem. Afterwards, realizing the desired ratio is $2$ should remind us of centroid length ratios. At this point, it's easy to see $\frac{BH}{HC} = 2$ if and only if $\frac{CH}{HM} = 2$, which motivates us to construct some triangle, i.e. $ACA_1$, where $CM$ is a median. The amount of intersections along with the harmonic quadrilateral $CDD_1F$ motivates us to use projective geometry, and the most natural approach is to show some cevian of $ACA_1$ containing $H$, i.e. $AG$, bisects the opposite side of the triangle. Also, $M \in D_1G$ follows from isogonality, as $DD_1 \parallel BC$ and $(D, T; C, B) = -1$. Edit: In my first solution, I could've used $$-1 = (C, D_1; D, F) \overset{B}{=} (C, K; D, G)$$to avoid projecting through $G$... oops.

Projective :3. Let $I$ the midppint of $BC$ and let $AI \cap CD=J$ and $BJ \cap AC=K$. By symetry we have that $K \in \omega$ meaning that $DKFC$ is harmonic. And now by ratio chasing: $$-1=(D, F; K, C) \overset{B}{=} (D, G; J, C) \overset{A}{=} (B, H; I, C) \implies BH=2HC$$Thus we are done

Same as #3, #11; posting for storage. Let $M$ be the midpoint of $\overline{BC},$ $X=\overline{AM}\cap\overline{CD},$ and $Y=\overline{AC}\cap\overline{BX}.$ Notice that $Y$ lies on $\omega$ as $BCYD$ is a isosceles trapezoid. Hence, $$-1=(DF;CY)\stackrel{B}=(DG;XC)\stackrel{A}=(BH;MC),$$which implies the desired result. $\square$