Prove that $\angle{MKN}$ is right angle.

Question

Two circles intersect at points $A$ and $B$. Line l passing through $A$ meets these circles in $C$ and $D$. Let $M$ and $N$ be midpoints of arcs $BC$ and $BD$ that do not contain $A$. Let $K$ be the midpoint of $CD$. Prove that $\angle{MKN}$ is right angle.

I tried some trigonometry but it went too complex.

This is true only if line $l$ doesn't go through area between two circles. — Alistair, Feb 18 '16 at 22:22
Here's a GeoGebra worksheet for this problem: https://www.geogebra.org/m/mvvhh3ds — Selrach Dunbar, Dec 12 '21 at 03:52

score 2 · Accepted Answer · answered Feb 18 '16 at 21:15

Let $C'$ be point such that $M$ is midpoint of $CC'$ and $D'$ such that $N$ is midpoint of $DD'$. Point $E$ is intersection of $CD'$ and $C'D$. Since $CM=MB=MC'$, $M$ is circumcenter so $CBC'$ is right triangle. Same goes for triangle $DD'B$. Because quadrilaterals $MBAC$ and $BNDA$ are cyclic, it follows that

$\angle BMC= 180^\circ-\angle BAC=\angle BAD = 180 ^\circ-\angle BND = \angle BND'$, and from here you can easily find that right triangles $CBC'$ and $DD'B$ are similar.

Triangles $CBD'$ and $C'BD$ are similar because $\angle CBD' = \angle C'BD $ and $\frac{BC}{BC'}= \frac{BD'}{BD}$ (That follows from similarity of right triangles $CBC'$ and $DD'B$.)

$\Rightarrow \angle BCE= \angle BC'E$ so quadrilateral $CBEC'$ is cyclic and $\angle CEC'= \angle CBC'=90^\circ$.

Since $DC' \parallel KM$ and $CD' \parallel KN$ (because $KM$ and $KN$ are midsegments in triangles $CC'D$ and $CD'D$) , $\angle MKN=\angle CED = 90^\circ$ .

Prove that $\angle{MKN}$ is right angle.

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