Angle bisectors $\overline{AX}$ and $\overline{BY}$ of triangle $ABC$ meet at point $I$. Find $\angle C,$ in degrees, if $\angle AIB = 109^\circ$. Could you please help. I just need a few hints. I don't want the answer.

LinTaeyang7 Jun 19, 2020

#1**+1 **

To make the calculations easier, let angle(B) = 2b and angle(A) = 2a.

To make the description easier, let point X be on BC and point Y be on AC.

In triangle(AXB), write angle(AXB) in terms of a and b. Then, write angle(CXI) in terms of a and b.

In triangle(AYB), write angle(AYB) in terms of a and b. Then, write angle(CYI) in terms of a and b.

Looking at triangle(AIB), how many degrees are in a + b?

How many degrees are in angle(XIY)?

Finally, look at quadrilateral(CXIY) --- you can get the answer!

geno3141 Jun 19, 2020