In triangle $ABC$, the bisector of angle $B$ meets the opposite side $AC$ at $B'$. Similarly, the bisector of angle $C$ meets the opposite side $AB$ at $C'$ . Prove that $A=60^{\circ}$ if, and only if, $BC'+CB'=BC$.
Source: Gulf Mathematical Olympiad 2013
Tags: trigonometry, geometry unsolved, geometry
In triangle $ABC$, the bisector of angle $B$ meets the opposite side $AC$ at $B'$. Similarly, the bisector of angle $C$ meets the opposite side $AB$ at $C'$ . Prove that $A=60^{\circ}$ if, and only if, $BC'+CB'=BC$.