8.5. Given a convex quadrilateral $A B C D$, where $A B=A D=1, \angle A=80^{\circ}$, $\angle C=140^{\circ}$. Find the length of the diagonal $A C$.
Answer: 1. Solution. We will prove that $A C=1$ by contradiction. If $A C>1$, then in triangle $A B C$ the larger side $A C$ is opposite the larger angle: $\angle B>\angle B C A$. Similarly, for triangle $A D C$ we have $\angle D>\angle D C A$. Adding these inequalities, we get $\angle B+\angle D>\angle C=140^{\circ}$. Thus, the sum of all angles of the quadrilateral $(\angle A+\angle C)+(\angle B+\angle D)>\left(80^{\circ}+140^{\circ}\right)+140^{\circ}=360^{\circ}$ - a contradiction. Similarly, if we assume that $A C<1$, then $\angle B+\angle D<\angle C$, and as a result, we get the sum of the angles of the quadrilateral less than $360^{\circ}$ - again, a contradiction. Comment. Another solution method uses the properties of inscribed angles: if we assume that point $C$ lies inside the unit circle centered at point $A$, then angle $C$ will be greater than $\frac{1}{2}\left(360^{\circ}-\angle A\right)=140^{\circ}$, and if $C$ lies outside the circle, then angle $C$ will be less than $140^{\circ}$. Therefore, $C$ lies on the circle.