In $\triangle A B C, \angle B=68^{\circ}$ and $\angle C=32^{\circ}$. Sides $\mathrm{AB}$ and $\mathrm{AC}$ are produced to points $\mathrm{D}$ and $\mathrm{E}$, respectively. The bisectors of $\angle D B C$ and $\angle B C E$ meet at $\mathrm{F}$. What is the measure of $\angle B F C$ ?

50°

\(\angle\)B = \({68}^\circ\) and \(\angle\)C = \({32}^\circ\)

\(\angle\)A = \({180}^\circ\) - \({68}^\circ\) - \({32}^\circ\) = \({80}^\circ\)

Now, \(\angle\)A = \({180}^\circ\) - \(\angle\)A/2 = \({90}^\circ\) - \({80}^\circ\)/2 = \({50}^\circ\).