My problem

Observe that $WY\perp XZ$ (this is one of the characterizations of a bicentric quadrilateral). Let $P=XY\cap WZ$ and $Q=WY\cap XZ$. Brocard then gives that $PQ$ is the polar of $K$ wrt the incircle $\Gamma$ of $ABCD$. After an inversion wrt $\Gamma$ we get that $A'$ is the midpoint of $WZ$, $C'$ is the midpoint of $XY$ and $K'$ is the foot of the perpendicular from $I$ to $PQ$ (because $PQ$ is the polar of $K$ wrt $\Gamma$). We have to prove that $A'C'\parallel IK'$, so we have to prove that $A'C'\perp PQ$. Since $WY\perp XZ$, Brahmagupta gives $A'Q\perp XY$ and $C'Q\perp WZ$, so $Q$ is the orthocentre of $A'C'P$, and therefore $PQ\perp A'C'$, as desired.

nao podia divulgar as questoes até dia 16

Unnecessary complex... Invert around $(I)$. $K^*$ is the altitude from $I$ to $BD$. We want to show that $A^*C^*\perp BD$ where $A^*,C^*$ are the midpoints of $WZ,XY$ respectively. By angle chasing we observe that $XZ\perp YW$. Let $(XYZW)$ be the unit circle. Note that $xz+yw=0$. \[\frac{\frac{x+y}{2}-\frac{z+w}{2}}{\frac{2xw}{x+w}-\frac{2yz}{y+z}}\overset{?}{=}-\frac{\frac{1}{2x}+\frac{1}{2y}-\frac{1}{2z}-\frac{1}{2w}}{\frac{2}{x+w}-\frac{2}{y+z}}\]\[\frac{(x-z)+(y-w)}{xyw+xzw-xyz-yzw}\overset{?}{=}\frac{xy(z+w)-zw(x+y)}{xyzw((y-w)-(x-z))}\iff \frac{xyzw((y-w)^2-(x-z)^2)}{(xyw-yzw)^2-(xyz-xwz)^2}\overset{?}{=}1\]\[-(xz)^2((y-w)^2-(x-z)^2)\overset{?}{=}(xz)^2(x-z)^2-(xz)^2(y-w)^2\]Which is true as desired.$\blacksquare$

Official Solution: Let \( P \) be the intersection of the diagonals of \( ABCD \). Lemma: The center of \( (AIC) \) is the intersection of the tangents to \( (ABC) \) at \( A \) and \( C \). Proof: Angle chasing. This means that \( (AIC) \) and \( (ABC) \) are orthogonal, so \( O \) is the intersection of the tangents to \( (AIC) \) at \( A \) and \( C \). Therefore, by the construction of the symmedian, we have that \( OI \) is the symmedian of \( AIC \). Since \( (KP;AC) \) is a harmonic bundle and \( O \), \( I \), and \( P \) are collinear (bicentric quadrilateral property), due to the symmedian we conclude that \( KI \) is tangent to \( (AIC) \).