Extend $ AL$ to meet $ BC$ at $ E$. Now $ \angle DLE = 180^{\circ}-\angle ALD = \angle ABC$. Since $ \angle DLB = \angle DBL$ (triangle $ DLB$ is isosceles) we must also have $ \angle ELB = \angle EBL$ so the triangle $ ELB$ is isosceles too. It means that $ ED$ is a perpendicular bisector of $ LB$ so $ \angle LED = \frac{1}{2}\angle LEB = 90^{\circ}-\angle LBE$. We also have $ \angle LEC = 2\angle LBE$ so $ \angle DEC = 90^{\circ}+\angle LBE$. Since $ \angle EAD = \angle DCE$ quadrilateral $ ADEC$ is cyclic so $ \angle DAC= 90^{\circ}-\angle LBE$. Note that $ \angle ACD = \angle LCE$ (from the conditions) and $ \angle ADC = \angle AEC$ (because $ ADEC$ is cyclic) so the triangles $ ADC$ and $ LEC$ are similar. Therefore $ LEC = \angle DAC = 90^{\circ}-\angle LBE$. So $ \angle BLC = \angle BLE+\angle ELC = 90^{\circ}$ and we are done.

Let $ AL$ meet $ BC$ at $ L_1$, $ \angle{DLL_1}=\angle{ABC}$ $ \Rightarrow$ $ LL_1=BL_1$, quadrilateral $ ADL_1C$ is cyclic, because $ \angle{DAL_1}=\angle{L_1CD}$ then, $ \angle{DL_1A}=\angle{DCA}$ $ \Rightarrow$ $ \angle{LCB}=\angle{DL_1A}=\angle{DL_1B}=90^o-\angle{LBL_1}$ then $ \angle{BLC}=90^o$

the main idea of this problem is prove ADIC is cyclic( many proof). Then <DIB=<A so <ALI= 180-2<A. then <LCI = <A(1) ,but we have <LCB = 90-<DIB=90-<A(2) From (1) and (2) we have <BLC= 90

Ya I was trying to follow the explanation but my geometry is not all up to date, is there like a online article or something I could read that would make the explanation more clear.

Sorry to revive it... a nice one indeed, Let $LD$ meet $CB$ again at $X$. Then just note that our angle condition gives $A,L,B,X$ cyclic and moreover a trapezoid. Now its finished by some angle chase...