KöMaL 2013 B4566

The squares ABDE , BCFG and CAHI are drawn over the sides of a triangle ABC , on the outside. The triangles DBG , FCI and HAE are completed to form the parallelograms DBGJ , FCIK and HAEL . Prove that $AKB\sphericalangle +BLC\sphericalangle + CJA\sphericalangle =90^{\circ}$.