Angle bisectors $\overline{AX}$ and $\overline{BY}$ of triangle $ABC$ meet at point $I$. Find $\angle C,$ in degrees, if $\angle AIB = 109^\circ$. [asy] pair A,B,C,X,Y,I; A= (0,0); B = (1,0); C = (0.6,0.5); X = intersectionpoint(B--C , A -- (bisectorpoint(B,A,C))); Y = intersectionpoint(A--C, B -- scale(6)*( (bisectorpoint(C,B,A)) - B)); I = intersectionpoint(A--X, B--Y); draw(X--A--C--B--Y); draw(A--B); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,N); label("$I$",I,S); label("$X$",X,NE); label("$Y$",Y,NW); [/asy]

Guest Jun 14, 2020

#2**+1 **

**Angle bisectors \(\overline{AX}\) and \(\overline{BY}\) of triangle \(ABC\) meet at point \(I\). Find \(\angle C\), in degrees, if \(\angle AIB = 109^\circ\).**

\(\begin{array}{|rcll|} \hline \text{In }\triangle AIB \\ \hline \mathbf{180^\circ} &=& \mathbf{109^\circ + A + B} \\ 180^\circ-109^\circ &=& A + B \\ 71^\circ &=& A + B \\ \mathbf{A+B} &=& \mathbf{71^\circ} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \text{In } AIXC \\ \hline \mathbf{360^\circ} &=& \mathbf{C + 109^\circ + 180^\circ-(C+B) + 180^\circ-(C+A)} \\ 0 &=& C + 109^\circ -(C+B) -(C+A) \\ 0 &=& C + 109^\circ -2C - (A+B) \\ 0 &=& -C + 109^\circ - (A+B) \\ C &=& 109^\circ - (A+B) \quad | \quad \mathbf{A+B=71^\circ} \\ C &=& 109^\circ - 71^\circ \\ \mathbf{C} &=& \mathbf{38^\circ} \\ \hline \end{array}\)

heureka Jun 15, 2020