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

<C =

Step-by-step explanation:

Given that: <AIB =

<AIB + <BIX = (sum of angles on a straight line)

+ <BIX =    

<BIX =   -

<BIX =

But,

<AIB = <YIX = (opposite angle property)

<XIB = <AIY =  (opposite angle property)

Therefore,

+ =   (Exterior angle property)

=  

A + B =

A + B + C =   (sum of angles in a triangle)

  + C =

C = -

C =

Thus, angle C is .

idk

step-by-step explanation:

i don't know