First round – City competition

1.5. It suffices to show that $\triangleq ECA=\triangleq HGA$. Since the quadrilateral $ACBE$ is cyclic, we have $\triangleq ECA=\triangleq EBA$, so it suffices to show that $ABGH$ is a cyclic quadrilateral. From triangle $DEH$ we have $\triangleq DHE=180^{\circ }-\triangleq EDH-\triangleq HED$, i.e. $$\triangleq AHB=\triangleq DHE=180^{\circ }-\triangleq FDA-(180^{\circ }-\triangleq CEB),$$ --- ## Croatia2016_booklet — Page 14 FINAL ROUND – NATIONAL COMPETITION so by using that $\angle CEB=\angle CAB$ (which holds because $ACBE$ is also cyclic) we get that $$\angle AHB=180^{\circ }-\angle FDA-(180^{\circ }-\angle CAB)=180^{\circ }-\angle FDA-\angle BAG.$$ Quadrilateral ADBF is cyclic, so we have $\angle FDA=\angle FBA=\angle GBA$. It follows that $$\angle AHB=180^{\circ }-\angle FDA-\angle BAG=180^{\circ }-\angle GBA-\angle BAG=\angle AGB.$$ Therefore, ABGH is a cyclic quadrilateral, which finishes the proof.