Angle bisectors AX and BY of triangle ABC meet at a point I. Find angle C in degrees, if angle AIB is 109 degrees.
This is the (really bad) diagram. https://latex.artofproblemsolving.com/9/a/4/9a410e2228e9f2f01c8a6c8990c2ea3c9ebff2a0.png
Thank you so much!!!!!
It's not a bad diagram. AoPS folks do a good job writing Asymptope code.
If you'll be so kind as to provide the week and problem number?
I agree the diagram is great, but the lighting on the problem is bad :). Sure, Week 7 Problem 7
So AIB is 109 degrees. IAB + IBA + 109 = 180, so IAB + IBA = 180 -109 = 71.
AX and BY are angle bisectors so IAB = CAB/2, and IBA = CBA/2. (CAB + CBA)/2 = 71.
Rearrange the last equation and you should be very close to your answer.
If you don't understand anything, feel free to ask!