Let $A(0;0), B(b;0), C(x_c;y_c), D(0;d)$. Denote with $M,N,P,Q$ the midpoints of $CB, BA, AD, DC$ respectively. Therefore, $M(\frac{x_c+b}{2}; \frac{y_c}{2}), N(\frac{b}{2};0), P(0; \frac{d}{2}), Q(\frac{x_c}{2}; \frac{y_c+d}{2}).$ Hence, $\overrightarrow{CD}(-x_c; d-y_c)$ and $\overrightarrow{CB}(b-x_c; -y_c).$ Since $\overrightarrow{CD}.\overrightarrow{CB}=0,$ we obtain $$x_c(x_c-b)+(y_c-d)y_c=0\iff x_c^2-bx_c+y_c^2-dy_c = 0. \hspace{2mm} (1)$$ Similarly, $\overrightarrow{MD}(-\frac{x_c+b}{2}; d-\frac{y_c}{2})$ and since $\overrightarrow{MN}=\frac{1}{2}\overrightarrow{CA}$, we obtain $\overrightarrow{MN}(-\frac{x_c}{2}; -\frac{y_c}{2}).$ Also, note that $\angle DMN = 90^{\circ}$, so $$\overrightarrow{MD}.\overrightarrow{MN}=\frac{x_c(x_c+b)}{4}+\frac{2d-y_c}{2}\cdot(-\frac{y_c}{2})=0,$$which is equivalent to $$x_c(x_c+b)-y_c(2d-y_c)=0 \iff x_c^2+bx_c-2dy_c+y_c^2=0. \hspace{2mm} (2)$$ It is enough to show that $\angle QPB=90^{\circ} \iff \overrightarrow{PQ}.\overrightarrow{PB}=0$. However, $\overrightarrow{PB}(b;-\frac{d}{2})$ and $\overrightarrow{PQ}(\frac{x_c}{2}; \frac{y_c}{2})$ because $\overrightarrow{PQ}=\frac{1}{2}\overrightarrow{AC}.$ Thus, $$\overrightarrow{PQ}.\overrightarrow{PB}=\frac{1}{2}x_cb-\frac{1}{4}y_cd \iff 4\overrightarrow{PQ}.\overrightarrow{PB}=2x_cb-y_cd.$$However, $$(2)-(1)=0=2bx_c-dy_c=4\overrightarrow{PQ}.\overrightarrow{PB},$$so $\overrightarrow{PQ}.\overrightarrow{PB}=0$ and we are done.

Also there is not hard complex bash. In fact, we need to prove that $DN \perp AC$ iff $BM \perp AC$, there $N, M$ are midpoints of $BC, AD$.

Denote by $N,P,R,M$ as the midpoints of $AB,AD,CD,BC$ respectively. By assumption $ANMD$ is cyclic, Thus $\angle BDC= \angle BAC= \angle BNM = \angle MDA$ which implies $\angle BDA= \angle MDC$. We have $\triangle BDA \sim \triangle MDC$ which then gives: $\frac{BA}{AD}=\frac{CM}{CD}$. Thus $\frac{AB}{AP}=\frac{BC}{CD}$ which gives $\triangle ABP \sim \triangle CBD$. Then $\angle PBC= \angle ABD= \angle ACD= \angle PRD$. Which gives $BCRP$ cyclic.