TUYMAADA 2007 Problem 6

Point $D$ is chosen on the side $AB$ of triangle $ABC$ . Point $L$ inside the triangle $ABC$ is such that $BD=LD$ and $\angle LAB=\angle LCA=\angle DCB$ . It is known that $\angle ALD+\angle ABC=180^\circ$ . Prove that $\angle BLC=90^\circ$ .