ATimo wrote: Suppose that $ABCD$ is a parallelogram such that $DAB>90$. Let the point $H$ to be on $AD$ such that $BH$ is perpendicular to $AD$. Let the point $M$ to be the midpoint of $AB$. Let the point $K$ to be the intersecting point of the line $DM$ with the circumcircle of $ADB$. Prove that $HKCD$ is concyclic. Let $\Gamma _1$ and $\Gamma _2$ denote the circles through $A,B,D$ and through $D,C,H$ respectively. Let $E$ be the intersection point of $HM$ and $BC$. Obviously, $E$ is on $\Gamma _2$. From $MA\cdot MB=MA^2=MH\cdot ME$ we have $MD$ is the radical axis of $\Gamma _1$ and $\Gamma _2$. Hence $K$ is on $\Gamma _1$ and radical axis of $\Gamma _1$ and $\Gamma _2$, so $K$ is on $\Gamma _2$.

Dear Mathlinkers, a variant 1. (M) the circle with diameter HE 2. according to the three chords theorem, we are done. Sincerely Jean-Louis

Let $ J $ the point of intersection of $ KC $ and $ AB $.Assume there $ AB=2 $ also that $ DM=k $ then $ KM = \frac{1}{k} $ locates a point $ L $ such that $ LJ $ is parallel to $ DB $ then $ \frac{KM}{KD}=\frac{KJ}{KC} $ also $ \frac{ML}{MJ}=\frac{LD}{JB} $ hence it is easy to see $ DL= \frac{2-2k}{k^2+1} $ also that $ DK=\frac{k^2+1}{k} $.By Stewart Theorem in the triangle $ DMH $ hence it is easy to see $ DA\cdot DH=k^2-1 $ then $ DA\cdot DH=DL\cdot DK=k^2-1$ then $ KJAH $ is cyclic as $ KLAJ $ is cyclic then $ HAJK $ is cyclic.$ AJ $ is parallel to $ BC $ then it is easy to see that $ DHBK $ is cyclic.

Reflect $H $ over $M $ to $H'$ $$AM \cdot BM=DM \cdot MK=HM \cdot MH' \implies DHKH' \text { is cyclic} $$$$\angle HDM=\angle MBK=\angle MH'K \implies MKH'B \text { is cyclic} $$Applying Reim Theorem gives $KH'CD $ is also cyclic $\implies $ $HDCK $ is cyclic