# problem about Circumscribed circle of triangle

Qeustion

circle O is Circumscribed circle of triangle $ABC$,where diameter $DE$ is perpendicular to segment $BC$ and intersects it in point $M$.perpendicular line from point $A$ to segment $DE$ intersects it on point $H$.perpendicular line from point $E$ to segment $AC$ intersects it on point $K$.prove $EK$ is tangent to Circumscribed circle of $HKM$ triangle. My IDEA

if i show that $\angle MKE$ is equal to half of arc $MK$,then i can concluded that $\angle MKE$ is Tangent Chord Angle So $EK$ is tangent.

quadrilateral $AHKE$ is Inscribed Quadrilateral because $\angle AHE = \angle AKE = 90$

$AHKE$ is Inscribed Quadrilateral so $\angle EAK = \angle EHK$

quadrilateral $MKCE$ is Inscribed Quadrilateral because $\angle EMC = \angle EKC = 90$

$MKCE$ is Inscribed Quadrilateral so $\angle MKE = \angle MCE$

And i stuck here.

## 1 Answer

well the problem was almost solved.$OM$ is perpendicular to $BC$ and $OB=OC=R$.so we can conclude that $O$ is a point on perpendicular bisector of $BC$.so $DE$ is perpendicular bisector of $BC$.Now we can easily prove that triangle $EMC$ is congruent with $EMB$. now we can says $\angle EBC = \angle BCE$,so arc $EC$ and $BE$ are equal.

and finally

$\angle EAK = \angle EHK = \frac {CE}{2} =\frac {BE}{2}=\angle MKE = \angle MCE$

so $\angle MHK = \angle MKE = \frac {MK}{2}$

now i showed that $\angle MKE = \frac {MK}{2}$.now i can conclude $\angle MKE$ is Tangent Chord Angle So $EK$ is tangent.