An angle inside an ISOSCELES triangle - Challenge

Question

I have a solution by Ceva's Theorem (https://youtu.be/AEmEneiEsuE) but I'd like to have a solution without trigonometry... Could it be possible?

Many thanks.

Answer 1

Let $AG$ be an altitude of $\Delta ABC,$ $I\in AG$ such that a ray $BD$ is a bisector of $\angle ABI$ and $CI\cap BD=\{F\}$.

Thus, since $$\measuredangle FIB=\measuredangle FIA=60^{\circ},$$ we obtain that a ray $AF$ is a bisector of $\angle BAI,$ which gives $$\measuredangle FAI=\frac{1}{2}\measuredangle BAG=10^{\circ}$$ and from here $$\measuredangle FAD=\measuredangle FAI+\measuredangle IAD=10^{\circ}+20^{\circ}=30^{\circ}$$ and since $$\measuredangle AFD=\measuredangle ABF+\measuredangle BAF=20^{\circ}+10^{\circ}=30^{\circ},$$ we obtain $$AD=FD,$$ which says $$F\equiv E$$ and $$\theta=50^{\circ}+30^{\circ}=80^{\circ}.$$