ABCD is a cyclic quadrilateral. $\mathrm{AB}$ and $\mathrm{DC}$ meet at $\mathrm{F}$, when produced. $\mathrm{AD}$ and $\mathrm{BC}$ meet at $\mathrm{E}$, when produced. If $\angle \mathrm{BAD}$ $=68^{\circ}$ and $\angle \mathrm{AEB}=27^{\circ}$, then what is the measure of $\angle \mathrm{BFC}$ ?

17°

\(\angle\)BAE + \(\angle\)BEA + \(\angle\)ABE = 180

= 68 + 27 + \(\angle\)ABE = 180

= 95 + \(\angle\)ABE = 180

= \(\angle\)ABE = (180 - 95)

= \(\angle\)ABE = \({85}^\circ\)

Now,

\(\angle\)ABE + \(\angle\)CBF = 180  (linear pair)

= 85 + \(\angle\)CBF = 180

= \(\angle\)CBF = (180 - 85)

= \(\angle\)CBF = 95

Now,

\(\angle\)BAE + \(\angle\)BCD = 180

= 68 + \(\angle\)BCD = 180

= \(\angle\)BCD = (180 - 68)

= \(\angle\)BCD = \({112}^\circ\)

Now,

\(\angle\)BCD + \(\angle\)BCF = 180

= 112 + \(\angle\)BCF = 180

= \(\angle\)BCF = (180 - 112)

= \(\angle\)BCF = \({68}^\circ\)

Now,

In \(\Delta \)BCF

\(\angle\)BCF + \(\angle\)CBF + \(\angle\)BFC = 180

= 68 + 95 +\(\angle\)BFC = 180

= 163 + \(\angle\)BFC = 180

= \(\angle\)BFC = (180 - 163)

= \(\angle\)BFC = \({17}^\circ\)

Therefore, answer is \({17}^\circ\).