$ABCD$ is a square. If $\angle EFB= \angle BFC$ what is $\angle EBF?$
I can only think of the Z-rule and say that $\angle BFC = \angle FBA$. After that I can't progress any further.
I think the problem might be missing some details and that it has infinitely many solutions, but moving the points $E$ and $F$ on the square (in GeoGebra) while maintaining the angle bisector, I only get $\angle EBF $ to be ~$45^o$
How can I solve this problem?
Let $K$ be the point on $EF$ such that $BK\perp EF$. Then $\triangle BCF\cong \triangle BKF$. This implies that $BK=BC=BA$ and hence $\triangle ABE\cong \triangle KBE$.
$\angle ABE=\angle KBE$ and $\angle CBF=\angle KBF$.
$\angle EBF=\angle ABC\div 2=45^\circ$.