circumcenter lies on circle, BK = AC, CL = AB (Kharkiv City XI 2016 - Ukr)

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circumcenter lies on circle, BK = AC, CL = AB (Kharkiv City XI 2016 - Ukr)

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#1 h Mar 18, 2020, 9:57 PM

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The circle $\omega$ passes through the vertices $B$ and $C$ of triangle $ABC$ and intersects its sides $AC,AB$ at points $A,E$ , respectively. On the ray $BD$ , a point $K$ such that $BK = AC$ is chosen , and on the ray $CE$ , a point $L$ such that $CL = AB$ is chosen. Prove that the center $O$ of the circumscribed circle of the triangle $AKL$ lies on the circle $\omega$ .

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#2 h Mar 19, 2020, 3:02 PM • 1 Y

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Let the circumcenter of $\triangle ALE$ and $\triangle ADK$ intersect at $A,O$
$\because \angle ACL=\angle ABK;CL=AB;BK=AC$
$\therefore \triangle ALC\cong \triangle KAB \Rightarrow\angle AOE=\pi -\angle ALE=\pi-\angle KAB$
$\therefore \angle EOD=2\pi -\angle AOE-\angle AOD=2\pi-(\pi-\angle KAB)-(\pi-\angle AKD)=\pi=\angle ABK=\angle EOD$
It means that point $O$ lies on $\omega$ .
And $\angle AOL=\angle AEL=\angle BEC=\angle BDC=\angle ADK=\angle AOK$
So the radium of $\odot ALO$ ia equal to the radium of $\odot AOK$
From the symmetry of the two circles, we know that $LO=OK$
And $\angle ALO=\angle AEO=\angle OCB=\pi-\angle OEB=\angle AED=\angle ALO$
$\therefore OL=OA$
So we are done

There are another two conclusions :
1. $O$ is the midpoint of arc $BEDC$
2.A fixed point is on the circumcircle of $\triangle ALK$

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This post has been edited 1 time. Last edited by ironball, Mar 19, 2020, 3:03 PM

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