find the angle in a triangle with angles $ 20^{\circ}, 70^\circ, 90^\circ $ I have triangle geometry problem:
a) Let $\triangle ABC$ be a right triangle with $\angle A=90^\circ$ and $\angle B=20^\circ$. Let BE be the angle bisector of $\angle B$, and $F$ be a point on segment $AB$ such that $\angle ACF=30^\circ$. prove that $\angle CFE=20^\circ$.
b) Prove the same statement with the condition $\angle ACF=40^\circ$ instead of $\angle ACF=30^\circ$.
 A: Here is a solution to part (b).
Let $\Omega$ be $BCE$'s circumcircle and $O$ its centre. Then $\angle COE=2\angle CBE=20^\circ$. $\angle EOB=2\angle ECB=140^\circ$ so $\angle COB=\angle COE+\angle EOB=160^\circ$ so $\angle OBC=\angle BCO=10^\circ$.
Erect an equilateral $\triangle BOJ$ on base $BO$.
$OJ=OB$ so $J$ is on $\Omega$. $\angle JCB=\frac{1}2 \angle JOB=30^\circ=\angle FCB$ so $F$ is on $CJ$.
Let $OJ$ cross $BF$ at $P$. $\angle OBJ=60^\circ=2\angle OBF$ so $BF$ is an axis of symmetry of $\triangle BOJ$ and is thus $OJ$'s $\perp$ bisector. Thus $\angle FOJ=\angle OJF$.
$OJ=OB=OC$ so $\triangle CJO$ is isosceles on base $CJ$. $\angle JCO=\angle FCO=\angle FCB+\angle BCO=40^\circ$, so $\angle OJC=\angle JCO=40^\circ$ so $\angle FOJ=\angle OJF=\angle OJC=40^\circ$.
Therefore $\angle EOF=\angle EOB-\angle FOJ-\angle JOB=140^\circ-40^\circ-60^\circ=40^\circ=\angle ECF$, so $OCEF$ is cyclic, so $\angle CFE=\angle COE=20^\circ$, which solves part b.

Hints for part (a): it might be useful to know that $OCEF_aF_b$ are all on a circle whose centre lies on $BC$.
