Focusing on Incenters #3

You have one degree of freedom here. So, you can arbitrarily set the value of $\angle A$ or $\angle B$ . Let, $\angle A=40^\circ$ . Then, you can easily get $\angle ADE=100^\circ; \angle AED=40^\circ$ . Now, consider the triangle ADE only. Imagine a bisector of $\angle A$ . In this bisector O is a point such that $\angle DOE=90^\circ+ \frac{\angle A}{2}=110^\circ$ . Hence, O is the incenter. Now, let $\angle ODE= \alpha$ . You have to find the $\alpha$ for which $\angle DOE=110^\circ$ . This can be a daunting task. I used coordinate geometry to find $\alpha=30^\circ$ . Finally, $\frac{\angle C}{2}=30^\circ+ \alpha$ $\angle C=120^\circ$ [P.S. Curious mind wants to know how the problem setter devised this ingenious problem!]