Inside triangle \( ABC \), a point \( D \) is chosen such that \( \angle ADB = \angle ADC \). The rays \( BD \) and \( CD \) intersect the circumcircle of triangle \( ABC \) at points \( E \) and \( F \), respectively. On segment \( EF \), points \( K \) and \( L \) are chosen such that \linebreak \( \angle AKD = 180^\circ - \angle ACB \) and \( \angle ALD = 180^\circ - \angle ABC \), with segments \( EL \) and \( FK \) \linebreak not intersecting line \( AD \). Prove that \( AK = AL \). Proposed by Matthew Kurskyi