$$ABCD$$ is a convex quadrilateral. Angles $$BAC$$, $$CAD$$, $$CBD$$, $$BDC$$ are $$50^\circ$$, $$60^\circ$$, $$30^\circ$$, $$25^\circ$$ respectively. If $$E$$ is the point of intersection of the line segments $$AC$$, $$BD$$ then angle $$AEB$$ (in degrees) is ________

In triangle $$BCD$$, $$\angle BCD = 180^\circ - 30^\circ - 25^\circ = 125^\circ$$, and in triangle $$ABD$$, $$\angle ABD + \angle ADB = 70^\circ$$. Let $$\angle ABD = \theta$$. Computing $$\frac{BE}{ED}$$ by the sine rule from triangle $$ABD$$ and from triangle $$CBD$$ and equating the two gives an equation in $$\theta$$ whose only solution is $$\theta = 35^\circ$$ (then $$\angle BCE = 65^\circ$$, $$\angle DCE = 60^\circ$$, and both ratios equal $$\frac{\sin 50^\circ}{\sin 60^\circ}$$). Hence $$\angle AEB = 180^\circ - 50^\circ - 35^\circ = 95^\circ$$.