Saudi Arabia Mathematical Competitions 2012

Let $I$ be the intersection point of the lines $BE$ and $CD$. The quadrilaterals $B{D}^{\prime }{B}^{\prime }D$ and $C{E}^{\prime }{C}^{\prime }E$ are cyclic, so $\overline{BD{B}^{\prime }}=\overline{B^{\prime }{D}^{\prime }I}$ and $\overline{CE{C}^{\prime }}=\overline{I{E}^{\prime }{C}^{\prime }}$. Since $BDEC$ is also cyclic, $\overline{BD{B}^{\prime }}=\overline{CE{C}^{\prime }}$. It follows that $\overline{B^{\prime }{D}^{\prime }I}=\overline{I{E}^{\prime }{C}^{\prime }}$, so $B^{\prime }{D}^{\prime }{E}^{\prime }{C}^{\prime }$ is a cyclic quadrilateral.

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Alternative solution.

Using the power of a point theorem, one has: $$I{B}^{\prime }\cdot ID=I{D}^{\prime }\cdot IB$$ $$I{C}^{\prime }\cdot IE=I{E}^{\prime }\cdot IC$$ $$IE\cdot IB=ID\cdot IC$$ From these one easily obtains $$I{B}^{\prime }\cdot I{E}^{\prime }=I{D}^{\prime }\cdot I{C}^{\prime }$$ which proves that the quadrilateral $B^{\prime }{D}^{\prime }{E}^{\prime }{C}^{\prime }$ is cyclic, using the reciprocal of the power of a point theorem.