68. National Mathematical Olympiad

It is well known fact that for cyclic hexagon the equation $|AB|\cdot |CD|\cdot |EF|=|BC|\cdot |DE|\cdot |AF|$ holds iff the diagonals $AD$, $BE$ and $CF$ are concurrent (follows from the trigonometric form of sine theorem for $△ACE$). Let $P$ be the point of intersection of $AD$, $BE$ and $CF$. Let us consider $△A{F}_{1}P$ and $△C{D}_{1}P$.



We have $△PA{F}_1=|△PAE-△F_{1}AE|=|△DAE-△FAE|=|△DCE-△FCE|=|△D_{1}CE-△PCE|=△PC{D}_1$. (Notice that $D_1$ and $D$ are on the opposite sides of $CF$ iff $F$ and $F_1$ are on the same side of $AD$). From $△APF\sim △CPD$, $|A{F}_1|=|AF|$ and $|C{D}_1|=|CD|$ we get: $$\frac{|A{F}_1|}{|C{D}_1|}=\frac{|AF|}{|CD|}=\frac{|AP|}{|CP|}$$ Therefore $△A{F}_{1}P\sim △C{D}_{1}P$. Hence $△AP{F}_1=△CP{D}_1$ and $\frac{|F_{1}P|}{|D_{1}P|}=\frac{|AP|}{|CP|}=\frac{|FP|}{|DP|}$ ($△APF\sim △CPD$). Since $D_1$ is outside of $△PCD$ iff $F_1$ is inside of $△FAP$, we have $△F_{1}P{D}_1=△APC=△FPD$. From $\frac{|F_{1}P|}{|D_{1}P|}=\frac{|FP|}{|DP|}$ and $△F_{1}P{D}_1=△FPD$ we have that $△F_1{D}_{1}P\sim △FDP$. Therefore $\frac{|F_1{D}_1|}{|FD|}=\frac{|F_{1}P|}{|FP|}=\frac{|D_{1}P|}{|DP|}$. Similarly $\frac{|D_1{B}_1|}{|DB|}=\frac{|D_{1}P|}{|DP|}=\frac{|F_1{D}_1|}{|FD|}$ and $\frac{|F_1{B}_1|}{|FB|}=\frac{|F_{1}P|}{|FP|}=\frac{|F_1{D}_1|}{|FD|}$. So $△F_1{D}_1{B}_1\sim △FDB$ and thus the solution is completed.