Angle bisectors $\overline{AX}$ and $\overline{BY}$ of triangle $ABC$ meet at point $I$. Find $\angle C,$ in degrees, if $\angle AIB = 109^\circ$.
In triangle(ABC), angle bisectors AX and BY intersect at point I.
-- angle(AIB) = 109o.
-- angle(ABI) + angle(BAI) = 180o - 109o = 71o
-- therefore, angle(A) + angle(B) = 2 · 71o = 142o
Since angle(A) + angle(B) + angle(C) = 180o, angle(C) = 180o - 142o = 38o.
