Školjka

Neka je $ABCD$ konveksan četverokut takav da je $|AB|\cdot |CD| = |BC|\cdot |DA|$. Točka $X$ leži unutar $ABCD$ tako da vrijedi \[\angle{XAB} = \angle{XCD}\quad\,\,\text{i}\quad\,\,\angle{XBC} = \angle{XDA}.\]Dokaži da je $\angle{BXA} + \angle{DXC} = 180^\circ$.

Let $ABC$ be an acute-angled triangle with $AB > AC$ and circumcircle $\Gamma$. Let $M$ be the midpoint of the shorter arc $BC$ of $\Gamma$, and let $D$ be the intersection of the rays $AC$ and $BM$. Let $E \neq C$ be the intersection of the internal bisector of the angle $ACB$ and the circumcircle of the triangle $BDC$. Let us assume that $E$ is inside the triangle $ABC$ and there is an intersection $N$ of the line $DE$ and the circle $\Gamma$ such that $E$ is the midpoint of the segment $DN$.\\ Show that $N$ is the midpoint of the segment $I_B I_C$, where $I_B$ and $I_C$ are the excentres of $ABC$ opposite to $B$ and $C$, respectively.