1
$\begingroup$

Question:
Circle $O$ is a circumscribed circle of triangle $ABC$,where diameter $DE$ is perpendicular to segment $BC$ and intersects it in point $M$.
A perpendicular line from point $A$ to segment $DE$ intersects it on point $H$.
A perpendicular line from point $E$ to segment $AC$ intersects it on point $K$.

Prove that $EK$ is tangent to Circumscribed circle of $HKM$ triangle. enter image description here

My IDEA:

If I show that $\angle MKE$ is equal to half of arc $MK$, then I can conclude that $\angle MKE$ is Tangent Chord Angle, therefore $EK$ is tangent. quadrilateral $AHKE$ is an inscribed quadrilateral because $\angle AHE = \angle AKE = 90$

$AHKE$ is an inscribed quadrilateral so $\angle EAK = \angle EHK$, and quadrilateral $MKCE$ is inscribed quadrilateral because $\angle EMC = \angle EKC = 90$. $MKCE$ is Inscribed Quadrilateral so $\angle MKE = \angle MCE.$

And I am stuck here.

$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.