## Balkan Mathematical Olympiad 2012 Problem 1

Let ${A}$, ${B}$ and ${C}$ be points lying on a circle ${\Gamma}$ with centre ${O}$. Assume that ${\angle ABC > 90}$. Let ${D}$ be the point of intersection of the line ${AB}$ with the line perpendicular to ${AC}$ at ${C}$. Let ${l}$ be the line through ${D}$ which is perpendicular to ${ A O }$. Let ${E}$ be the point of intersection of ${l}$ with the line ${AC}$, and let ${F}$ be the point of intersection of ${\Gamma}$ with ${l}$ that lies between ${D}$ and ${E}$. Prove that the circumcircles of triangles ${BFE}$ and ${CFD}$ are tangent at ${F}$.

Balkan Mathematical Olympiad 2012 Problem 1

Solution: Note that the circles are tangent if $\angle CDF +\angle EBF=\angle CFE$, and this condition is equivalent to $\angle FCD=\angle FBE$. Since $E$ is the orthocenter of the triangle formed by $AD, l, DC$ it follows that $BE \perp AD$, and the desired condition is equivalent to $\angle FCA=\angle FBD$, which is true because $ABFC$ is cyclic.