In the acute triangle $ABC$ the side $BC> AB$, and the angle bisector $BL = AB$. On the segment $BL$ there is a point $M$, for which $\angle AML = \angle BCA$. Prove that $AM = LC$.
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Tags: geometry, equal segments, angle bisector, equal angles
In the acute triangle $ABC$ the side $BC> AB$, and the angle bisector $BL = AB$. On the segment $BL$ there is a point $M$, for which $\angle AML = \angle BCA$. Prove that $AM = LC$.