Angle bisectors \(\overline{AX}\) and \(\overline{BY}\) of triangle $ABC$ meet at point $I$. Find \(\angle C\), in degrees, if \(\angle AIB = 109^\circ\).
+26367 Angle bisectors \(\overline{AX}\) and \(\overline{BY}\) of triangle \(ABC\) meet at point \(I\).
Find \(\angle C\), in degrees, if \(\angle AIB = 109^\circ\).
\(\begin{array}{|rcll|} \hline \text{In }\angle ABC \\ \hline 2x+2y +C &=& 180^\circ \\ 2x+2y &=& 180^\circ-C \\ 2(x+y) &=& 180^\circ-C \\ \mathbf{x+y} &=& \mathbf{\dfrac{180^\circ-C}{2}} \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline \text{In } IXCY \\ \hline 360^\circ &=& 109^\circ + \Big(180^\circ-(C+x)\Big) + C + \Big(180^\circ-(C+y)\Big) \\ 360^\circ &=& 109^\circ + 180^\circ-C-x + C + 180^\circ-C-y \\ 0 &=& 109^\circ -C-x + C -C-y \\ 0 &=& 109^\circ -C-x -y \\ C &=& 109^\circ -x -y \\ C &=& 109^\circ -(x+y) \quad | \quad \mathbf{x+y=\dfrac{180^\circ-C}{2}} \\ C &=& 109^\circ - \left(\dfrac{180^\circ-C}{2}\right) \quad | \quad * 2 \\ 2C &=& 218^\circ - (180^\circ-C) \\ 2C &=& 218^\circ - 180^\circ + C \quad | \quad -C \\ C &=& 218^\circ - 180^\circ \\ \mathbf{C} &=& \mathbf{38^\circ} \\ \hline \end{array}\)
