IMO 1962 - Problem 5

Problem: On the circle $k$ three point $A,B$ and $C$ are given. Construct the fourth point on the circle $D$ such that one can inscribe in a circle in $ABCD$.
Solution: Analysis. Let $ABCD$ be the desired quadrilateral. Let us assume w.l.o.g that $AB>BC$( for $AB=BC$ the construction is trivial). For a tangent quadrilateral we have $AD-DC=AB-BC$ and $\angle{AXC}=\angle{ADC}+\angle{CDX}=180^0-\frac{\angle{ABC}}{2}$. Constructing $X$ and hence $D$ is how obvious.