Triangle Bisector Challenge

Question

I've found this olympic geometry problem, which remains to be unsolvable by all my teachers and friends;

In a triangle $ABC$ $[BE],[CD]$ are bisectors, $[BE]\cap[CD]=F$

$m(\widehat{FDE})=18^\circ$ $m(\widehat{FED})=24^\circ$

According to the given knowledge find $m(\widehat{EBC})=\alpha$ (m: the measure of the angle) (sometimes used differently in other countries)

My Attempts

This problem has a complete smackdown on me, I have tried to take the symmetric of point $D$ (in respect to line $BC$) which I named $D'$ then drew an edge between $D$ and $D'$ $DBD'C$ is also a deltoid....(1)

I also was able to find that $m(\widehat{A})=96^\circ$...(3)

And that $\alpha+<FCB=42^\circ$

I am now stuck here, a full solution might be incredible but I am open to any sorts of direction.

Thank you:))

Answer 1

As you noticed, $\widehat{EFD}=\widehat{BFC}=\pi-\frac{B}{2}-\frac{C}{2}=90^\circ+\frac{A}{2}=138^\circ$ from which $A=96^\circ$.
By the sine theorem $$ \frac{\sin 18^\circ}{CE}=\frac{\sin(C/2)}{ED},\qquad \frac{\sin 24^\circ}{BD}=\frac{\sin(B/2)}{ED}, $$ hence $$\frac{\sin(B/2)}{\sin(C/2)}=\frac{\sin(24^\circ)}{\sin(18^\circ)}\cdot\frac{CE}{BD}=\frac{\sin(24^\circ)}{\sin(18^\circ)}\cdot\frac{\frac{ab}{a+c}}{\frac{ac}{a+b}}=\frac{\sin(24^\circ)}{\sin(18^\circ)}\cdot\frac{\sin B}{\sin C}\cdot\frac{\sin(A)+\sin(B)}{\sin(A)+\sin(C)}$$ leads to a trigonometric equation in $\sin(\alpha)=\sin(B/2)$, since $\frac{C}{2}=42^\circ-\frac{B}{2}$ and $\sin(A)$ is known.

Answer 2

The hint.

We have $$\frac{\beta}{2}+\frac{\gamma}{2}=18^{\circ}+24^{\circ}$$ or $$\beta+\gamma=84^{\circ}.$$ In another hand, by law of sines we obtain: $$\frac{DF}{EF}=\frac{\sin24^{\circ}}{\sin18^{\circ}}.$$ from this system we can find $\frac{\beta}{2}$.

I got $\frac{\beta}{2}=6^{\circ}$.