Let $L$ be the other intersection of $BM$ with the circumcircle of $ABCD$, with a simple angle chasing you get $\angle ABL= \angle BAC$ then $AM=MC=MB$ and $\angle ABC=\angle ADC= 90$

mathisreal wrote: Let $ABCD$ be a cyclic quadrilateral such that the lines $AB$ and $CD$ intersects in $K$, let $M$ and $N$ be the midpoints of $AC$ and $CK$ respectively. Find the possible value(s) of $\angle ADC$ if the quadrilateral $MBND$ is cyclic. Solved with frost456 $\color{blue}\boxed{\textbf{Answer: 90°}}$ $\color{blue}\boxed{\textbf{Proof:}}$ $\color{blue}\rule{24cm}{0.3pt}$ Working in complex numbers, where $\odot (ABCD)$ is the unit circle, let $a=a$, $b=b$, $c=1$ and $d=d$ $K,C$ and $D$ are collinears: $$\Rightarrow \frac{k-1}{\overline{k}-1}=\frac{d-1}{\overline{d}-1}=-d$$$$\Rightarrow \overline{k}=\frac{d+1-k}{d}...(I)$$$K,B$ and $A$ are collinears: $$\Rightarrow \frac{k-b}{\overline{k}-\overline{b}}=\frac{a-b}{\overline{a}-\overline{b}}=-ab$$$$\Rightarrow \overline{k}=\frac{a+b-k}{ab}...(II)$$By $(I)$ and $(II):$ $$\Rightarrow \frac{d+1-k}{d}=\frac{a+b-k}{ab}$$$$\Rightarrow k=\frac{abd+ab-ad-bd}{ab-d}...(\blacktriangle)$$$M$ is the midpoint of $AC:$ $$\Rightarrow m=\frac{a+1}{2}...(\blacksquare)$$$N$ is the midpoint of $CK:$ $$\Rightarrow n=\frac{k+1}{2}...(\star)$$$MBND$ is cyclic: $$\Rightarrow \frac{(m-n)(b-d)}{(b-n)(d-m)}\in\mathbb{R}$$By $(\blacksquare)$ and $(\star):$ $$\Rightarrow \frac{(\frac{a-k}{2})(b-d)}{(\frac{2b-k-1}{2})(\frac{2d-a-1}{2})}\in\mathbb{R}$$$$\Rightarrow \frac{(a-k)(b-d)}{(2b-k-1)(2d-a-1)}\in\mathbb{R}$$By $(\blacktriangle):$ $$\Rightarrow \frac{(\frac{a^2b-ad-abd-ab+ad+bd}{ab-d})(b-d)}{(\frac{2ab^2-2bd-abd-ab+ad+bd-ab+d}{ab-d})(2d-a-1)}\in\mathbb{R}$$$$\Rightarrow \frac{b(a-1)(a-d)(b-d)}{(b-1)(2ab-ad-d)(2d-a-1)}\in\mathbb{R}$$$$\Rightarrow \frac{b(a-1)(a-d)(b-d)}{(b-1)(2ab-ad-d)(2d-a-1)}=\frac{1}{(\overline{\frac{b(a-1)(a-d)(b-d)}{(b-1)(2ab-ad-d)(2d-a-1)}})}$$$$\Rightarrow \frac{b(a-1)(a-d)(b-d)}{(b-1)(2ab-ad-d)(2d-a-1)}=\frac{(\frac{1}{b})(\frac{1-a}{a})(\frac{d-a}{ad})(\frac{d-b}{bd})}{(\frac{1-b}{b})(\frac{2a-d-ad}{ad})(\frac{2d-b-ab}{abd})}$$$$\Rightarrow \frac{b}{(2d-a-1)(2ab-ad-d)}=\frac{1}{(2a-d-ad)(2d-b-ab)}$$$$\Rightarrow b(2a-d-ad)(2d-b-ab)=(2d-a-1)(2ab-ad-d)$$$$\Rightarrow b(a+1)(2ab+2d^2-bd-abd)=(a+1)(2d^2+2ab-ad-d)$$$\color{red}\boxed{\textbf{If } a\neq -1}$ $\color{red}\rule{24cm}{0.3pt}$ $$\Rightarrow b(2ab+2d^2-bd-abd)=2d^2+2ab-ad-d$$$$\Rightarrow (b-1)(2ab+2d^2-bd-d-abd-ad)=0$$$$\Rightarrow 2ab+2d^2-bd-d-abd-ad=0$$$$\Rightarrow a=\frac{bd+d-2d^2}{2b-bd-d}=\frac{(b+1-2d)d}{2b-bd-d}$$$$\Rightarrow \frac{(b+1-2d)d}{2b-bd-d}=\frac{1}{(\overline{\frac{(b+1-2d)d}{2b-bd-d}})}$$$$\Rightarrow \frac{(b+1-2d)d}{2b-bd-d}=\frac{\frac{2d-1-b}{bd}}{(\frac{d+bd-2b}{bd})(\frac{1}{d})}$$$$\Rightarrow d+bd-2b=bd-2b-d$$$$\Rightarrow d=0(\Rightarrow \Leftarrow)$$$\color{red}\rule{24cm}{0.3pt}$ $$\Rightarrow a=-1$$$$\Rightarrow AC \text{ is diameter}$$$$\Rightarrow \boxed{\angle ADC=90°}_\blacksquare$$$\color{blue}\rule{24cm}{0.3pt}$