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Intersection of circumcircles of MNP and BOC
Djile   39
N an hour ago by bjump
Source: Serbian National Olympiad 2013, Problem 3
Let $M$, $N$ and $P$ be midpoints of sides $BC, AC$ and $AB$, respectively, and let $O$ be circumcenter of acute-angled triangle $ABC$. Circumcircles of triangles $BOC$ and $MNP$ intersect at two different points $X$ and $Y$ inside of triangle $ABC$. Prove that \[\angle BAX=\angle CAY.\]
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Djile
Apr 8, 2013
bjump
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Prove a geometry ratio
nhhlqd   1
N Mar 29, 2021 by PROF65
Given triangle $ABC$ with $AB<AC$ inscribed in a circle centered $O$, denoted as $\Gamma$. Let $I$ be an arbitrary point on the circumcircle of triangle $BOC$. Given $E$, $F$ be the moving points on $\Gamma$ such that $BE\parallel CF$. Suppose $BI\cap AF = P$; $CI\cap AE = Q$ and $EF\cap PQ = K$. Shown that the ratio $\dfrac{KE}{KF}$ is not change as $E$ and $F$ move.

This problem created when I tried to generalize a problem. However, I couldn't solve this problem with the way I've done with the previous one.
Any help is appreciated.
Thanks a lot.
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nhhlqd
Mar 29, 2021
PROF65
Mar 29, 2021
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Prove a geometry ratio
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nhhlqd
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nhhlqd
#1 h  Mar 29, 2021, 8:45 AM • 1 Y
Y by Mango247
Given triangle $ABC$ with $AB<AC$ inscribed in a circle centered $O$, denoted as $\Gamma$. Let $I$ be an arbitrary point on the circumcircle of triangle $BOC$. Given $E$, $F$ be the moving points on $\Gamma$ such that $BE\parallel CF$. Suppose $BI\cap AF = P$; $CI\cap AE = Q$ and $EF\cap PQ = K$. Shown that the ratio $\dfrac{KE}{KF}$ is not change as $E$ and $F$ move.

This problem created when I tried to generalize a problem. However, I couldn't solve this problem with the way I've done with the previous one.
Any help is appreciated.
Thanks a lot.
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PROF65
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PROF65
#2 h  Mar 29, 2021, 1:17 PM
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Let $BF \cap CE= K$ since $BE\parallel CF$ then $ \angle BKE=2 \angle BAC$ i.e.
$K\in (BOC) \implies \angle ICE= \angle IBF$ ,we have also $BF=CE $ because $BCFE$ is trapezoid $ (^*)$.
By Menelaus's it suffices to show that $\frac{PF}{PA}.\frac{QA}{AE} $ is constant .

$\frac{PF}{PA}.\frac{QA}{AE} =\frac{BF}{BA}.\frac{\sin \angle IBF}{\sin\angle IBA}.\frac{CA}{CE}.\frac{\sin \angle ICA}{\sin \angle ICE} \overset{(^*)}{=} $

$\frac{CA}{BA}.\frac{\sin \angle ICA}{\sin \angle IBA} $ which obviously depends only on the choice of $I$
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