Two sine theorems for triangles $ABC$ and $BLC$ gives us that angles $LBC$ and $ACB$ are equals, from where immediately $BK=LC=AB$

From $\sphericalangle BAC + \sphericalangle BLC$, we have $\sin\sphericalangle BAC =\sin \sphericalangle BLC$. Sine theorem for $\Delta BLC$ and $\Delta ABC$ get $$\frac{LC}{\sin \sphericalangle LBC}=\frac{CB}{\sin \sphericalangle BLC} \text{ and } \frac{AB}{\sin \sphericalangle BCA}=\frac{CB}{\sin \sphericalangle BAC}\text{.}$$From this we get $\sin \sphericalangle LBC=\sin \sphericalangle BCA$. Since $L$ is in $\Delta ABC$, it is clear that $\sphericalangle LBC+\sphericalangle BCA=180^{\circ}$ is impossible. Therefore, $\sphericalangle LBC=\sphericalangle BCA$, so the trapezoid $CKLB$ is isosceles. It's diagonals $BK$ and $AC$ are equal, hence $BK=LC=AB$.

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Let $ABDC$ be an isosceles trapezoid, i.e. cyclic, with $BD\parallel AC$ and $L$ reflection of $D$ in $BC$; then $\widehat{BLC}+\widehat{BAC}=180^\circ$ and $CL=CD=AB$. Then $\widehat{CBL}=\widehat{CBD}=\widehat{ACB}=\widehat{AKL}$ and $BCKL$ is cyclic, hence isosceles trapezoid, thus $BK=CL$, but $CL=AB$, done. Best regards, sunken rock

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Let T be a point such that BLCT is parallelogram, then ∠BLC=∠BTC, which means that ABTC is cyclic. And we have AB=CL=BT which means that ∠LBC=∠BCT=∠ACB, then BLKC is isosceles trapezoid, then BK=LC=BT=AB, which completes the solution.