minecraftfaq wrote: Inscribed Triangle $ABC$ on circle $\odot O$. Bisector of $\angle ABC$ intersects $\odot O$ at $D$. Two lines $PB$ and $PC$ that are tangent to $\odot O$ intersect at $P$. $PD$ intersects $AC$ at $E$, $\odot O$ at $F$. Prove that $M,F,C,E$ are concyclic. whats $M$? is it midpoint of $BC$?if that this is too easy...

I'm assuming $M$ is the midpoint of $BC$, let the tangents to $(ABC)$ at $D$ and $F$ intersect at $Q$ and let $G= PD \cap BC$, since $BDCF$ is an harmonic quadrilateral we get that $Q$ lies on $BC$, furthermore, we get $(Q, G, C, B) =-1$, but $M$ is midpoint of $BC$ and it is very well known that $QC \cdot QB= QG \cdot QM$ but $QC \cdot QB=QD^2$ so $QG \cdot QM= QD^2$ $\implies$ $\angle GDQ=\angle GFQ=\angle GMD$, hence $DMFQ$ is cyclic. Now $\angle CBD=\angle CDQ$ and $\angle ABD=\angle ACD$ so $AC \| DQ$. Now angle chase to get $\angle MFD=\angle MQD=\angle MCA$ and the conclusion follows.

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Hax.Mode.ON3 wrote: minecraftfaq wrote: Inscribed Triangle $ABC$ on circle $\odot O$. Bisector of $\angle ABC$ intersects $\odot O$ at $D$. Two lines $PB$ and $PC$ that are tangent to $\odot O$ intersect at $P$. $PD$ intersects $AC$ at $E$, $\odot O$ at $F$. Prove that $M,F,C,E$ are concyclic. whats $M$? is it midpoint of $BC$?if that this is too easy... Ricochet wrote: I'm assuming $M$ is the midpoint of $BC$, let the tangents to $(ABC)$ at $D$ and $F$ intersect at $Q$ and let $G= PD \cap BC$, since $BDCF$ is an harmonic quadrilateral we get that $Q$ lies on $BC$, furthermore, we get $(Q, G, C, B) =-1$, but $M$ is midpoint of $BC$ and it is very well known that $QC \cdot QB= QG \cdot QM$ but $QC \cdot QB=QD^2$ so $QG \cdot QM= QD^2$ $\implies$ $\angle GDQ=\angle GFQ=\angle GMD$, hence $DMFQ$ is cyclic. Now $\angle CBD=\angle CDQ$ and $\angle ABD=\angle ACD$ so $AC \| DQ$. Now angle chase to get $\angle MFD=\angle MQD=\angle MCA$ and the conclusion follows. Yes! $M$ is the midpoint of $BC$. Fixed.

Well,I first used complex numbers to show $ME||DF$, then we are done, Didnt't have to think anything for this problem.

$DBFC$ is a harmonic quadrilateral $\Rightarrow \angle PMC =\angle DMC =\angle BDM +\angle DBC=\angle DAC +\angle PDC=\angle PEC$. $\Rightarrow M,F,C,E$ are concyclic.

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[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -33.53903948671562, xmax = 33.10260478911287, ymin = -21.36281432153444, ymax = 18.482972986989544; /* image dimensions */ /* draw figures */ draw(circle((-1.982802344321283,1.4174368792414813), 10.275268604435189), linewidth(0.4) + red); draw((-10.433445211217208,7.262758087650175)--(-10.48,-4.36), linewidth(0.4) + blue); draw((5.307900100635554,-5.823198642064079)--(-10.48,-4.36), linewidth(0.4) + blue); draw((5.307900100635554,-5.823198642064079)--(-10.433445211217208,7.262758087650175), linewidth(0.4) + blue); draw((-3.4733101091557,-14.66512840883977)--(-10.48,-4.36), linewidth(0.4) + blue); draw((-3.4733101091557,-14.66512840883977)--(5.307900100635554,-5.823198642064079), linewidth(0.4) + blue); draw((-3.4733101091557,-14.66512840883977)--(4.585830034538148,9.318969234721077), linewidth(0.4) + blue); draw(circle((1.3093518530036736,-6.013873585107586), 4.003091932812015), linewidth(0.4) + linetype("4 4")); draw((-2.586049949682223,-5.09159932103204)--(0.7663741535233614,-2.0477772372597816), linewidth(0.4) + blue); draw((-1.5184133989490065,-8.847332372948628)--(5.307900100635554,-5.823198642064079), linewidth(0.4) + blue); draw((-10.48,-4.36)--(4.585830034538148,9.318969234721077), linewidth(0.4) + blue); draw((-10.48,-4.36)--(-1.5184133989490065,-8.847332372948628), linewidth(0.4) + blue); draw((4.585830034538148,9.318969234721077)--(5.307900100635554,-5.823198642064079), linewidth(0.4) + blue); draw((-10.433445211217208,7.262758087650175)--(-1.5184133989490065,-8.847332372948628), linewidth(0.4) + blue); draw((-10.433445211217208,7.262758087650175)--(4.585830034538148,9.318969234721077), linewidth(0.4) + blue); draw((-1.5184133989490065,-8.847332372948628)--(-2.586049949682223,-5.09159932103204), linewidth(0.4) + blue); /* dots and labels */ dot((-10.433445211217208,7.262758087650175),dotstyle); label("$A$", (-10.266663711540527,7.695030571581304), NE * labelscalefactor); dot((-10.48,-4.36),dotstyle); label("$B$", (-10.310163479344594,-3.91940743210418), NE * labelscalefactor); dot((5.307900100635554,-5.823198642064079),dotstyle); label("$C$", (5.480252233531218,-5.3983995374424065), NE * labelscalefactor); dot((-1.982802344321283,1.4174368792414813),linewidth(4pt) + dotstyle); label("$O$", (-1.8277087575518025,1.7790621502283983), NE * labelscalefactor); dot((-2.586049949682223,-5.09159932103204),linewidth(4pt) + dotstyle); label("$M$", (-2.3932057390046553,-4.745903020381424), NE * labelscalefactor); dot((4.585830034538148,9.318969234721077),linewidth(4pt) + dotstyle); label("$D$", (4.740756180862102,9.652520122764251), NE * labelscalefactor); dot((-3.4733101091557,-14.66512840883977),linewidth(4pt) + dotstyle); label("$P$", (-3.3067008628900325,-14.31585193727583), NE * labelscalefactor); dot((-1.5184133989490065,-8.847332372948628),linewidth(4pt) + dotstyle); label("$E$", (-1.349211311707081,-8.486883051531056), NE * labelscalefactor); dot((0.7663741535233614,-2.0477772372597816),linewidth(4pt) + dotstyle); label("$F$", (0.9562763819083955,-1.7009192740968404), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] We have that $ED$ is the $E$-symmedian of $EBC$, thus we have that $\angle MEC = \angle DEB = 180 - \angle BAD = \gamma + \frac{1}{2}\beta$. Since we have that $\angle MEC = \angle MEF + \angle FEC = \angle MEF + \frac{1}{2}\beta$, out of here we have that $\angle MEF = \gamma = \angle MCF$, thus we have that $MFCE$ is cyclic.

Since $BCFD$ is an harmonic quadrilateral let the tangents from $D,F$ and $BC$ intersect at $G$. By angle chasing $$\angle DGC=180-\angle CDG=\angle DCG=180-\angle BAC-\frac{\angle ABC}{2}-\frac{\angle ABC}{2}=180-\angle BAC-\angle ABC=\angle ACB \implies DG \parallel AC$$By the symedian $\triangle DMC \sim \triangle DBF$ which means that $\angle DBF=\angle DMC=\angle FDG$ hence $FMDG$ is cyclic and now since $DG \parallel AC$ we have $FMEC$ cyclic as desired, thus we are done