In $\Delta ABC$, $K$ and $L$ are points on $BC$. $K$ is closer to $B$ than $L$. $AL$ is the bisector of $\angle KAC$. $KL\times BC=BK\times CL$. Find $\angle BAL$.
I have tried using angle-bisector theorem on $\Delta AKC$ without any result. I feel the angle is a right angle and have even tried to apply converse of Pythagoras' theorem on $\Delta ABL$. I tried applying Stewart's theorem on $\Delta ABL$ but I couldn't get the form of Pythagoras' theorem.
Thanks for any help.
On
By the bisector theorem we have
$$ \frac{AC}{AK}=\frac{CL}{KL}=\frac{BC}{BK} $$
hence
$$\mu= \frac{AC}{CB}=\frac{AK}{KB} $$
and $K$ belongs to the Apollonius circle that is the locus of points $P$ such that $\frac{PA}{PB}=\lambda$.
Such a circle goes through the feet of the angle bisectors (the internal and the external) from $C$.
If $ABC$ is an acute-angled triangle, such a circle intersects the $BC$ segment only at $C$, hence there is no solution.
On the other hand, if we allow $\widehat{BAC}$ to be greater than $90^\circ$, we have a solution at $\widehat{BAL}=\color{red}{90^\circ}$:
The identity $KL \cdot BC = BK \cdot CL$ implies that line $AB$ is the external bisector of triangle $CKA$ at $A$ while $AL$ is the interior bisector at $A$. Hence they are orthogonal and thus the angle $\angle \, BAL = 90^{\circ}$.