Let $ABC$ be a triangle with $\angle A=90^\circ$ and $\angle B =20^\circ$. Let $E$ and $F$ be points on $AC$ and $AB$ respectively such that $\angle ABE = 10^\circ$ and $\angle ACF = 30^\circ$. Determine $\angle CFE$.
The answer is $20^\circ$ but it doesn't show in my solution:
Let the intersection of $EB$ and $FC$ be $H$. $BH$ is an angle bisector.
$$\frac{BF}{BC} = \frac{FH}{CH}$$
Let $\angle CFE$ be $a$. then $\angle EHF$ is $50^\circ-a$ (angle chasing).
Using Sine Law on triangles $EFH$ and $EHC$ and $BFC$, we have,
$$\frac{EH}{FH} = \frac{\sin(a)}{\sin(50^\circ-a)},$$
$$\frac{EH}{CH} =\frac{\sin(30^\circ)}{\sin(100^\circ)},$$
$$\frac{BF}{BC}=\frac{\sin(40^\circ)}{\sin(100^\circ)}.$$
Using the 4 equations, I got $$\sin(100^\circ)\sin(a)\sin(40^\circ)= \sin(30^\circ)\sin(120^\circ)\sin(50^\circ-a)$$ but obviously $a$ is not $20^\circ$.
Can someone spot my mistake? I can't seem to pinpoint it.
Note that $FC = 2AF$. Let $D$ be the midpoint of $BC$ and let $G$ be the point on $AB$ such that $GD$ is perpendicular to $BC$. Then triangles $ABC$ and $DBG$ are similar, so that $\dfrac{BD}{BG}=\dfrac{BA}{BC}$. Then by symmetry we have $\angle GCB=\angle GBC=20^{\circ}$, so that $\angle GCF=20^{\circ}$ also. Hence $CG$ bisects $\angle BCF$ so that $\dfrac{FC}{FG}=\dfrac{BC}{BG}$. Since $BE$ bisects $\angle ABC$, $\dfrac{BA}{BC}=\dfrac{AE}{CE}$. Now $$\frac{AF}{FG}=\frac{\dfrac12FC}{FG}=\frac{\dfrac12 BC }{BG}=\frac{BD}{BG}=\frac{BA}{BC}=\frac{AE}{EC}$$
It follows that $CG$ is parallel to $EF$, so that $\angle CFE=\angle GCF=20^{\circ}$