Midpoint of bisector; prove that B, E, F, C cyclic

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Midpoint of bisector; prove that B, E, F, C cyclic

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Source: Taiwan 2014 TST1, Problem 4

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6857 posts

#1 h Jul 18, 2014, 7:46 PM • 6 Y

Y by Davi-8191, A-Thought-Of-God, HamstPan38825, Hasin_Ahmad, Adventure10, Mango247

Let $ABC$ be an acute triangle and let $D$ be the foot of the $A$ -bisector. Moreover, let $M$ be the midpoint of $AD$ . The circle $\omega_1$ with diameter $AC$ meets $BM$ at $E$ , while the circle $\omega_2$ with diameter $AB$ meets $CM$ at $F$ . Assume that $E$ and $F$ lie inside $ABC$ . Prove that $B$ , $E$ , $F$ , $C$ are concyclic.

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#2 h Jul 18, 2014, 8:11 PM • 1 Y

Y by Adventure10

Take a look at this

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Promi

#3 h Jul 30, 2019, 5:19 PM • 5 Y

Y by PNT, Adventure10, Mango247, Mango247, Mango247

Let $N$ be the intersection of $CM$ and the perpendicular line through $A$ on $AM$
Define $J=CN\cap AB$
So, $(N,M;J,C)=-1$
So, taking the pencil from point $B$ we get, $BN||AM$
$N$ lies on the circle with diameter $AB$
Consequently, $BN||AM||CH$ , where $H=NA\cap BM$
$H$ lies on the circle with diameter $CH$
Let $K$ be the second intersection of the two circles, which lies on $BC$
Then, $\angle{KFM}=\angle{KAN}$
And $\angle{KEM}=\angle{KAH}$
So, $\angle{KEM}=\angle{KFM}$
So, $KEMF$ is cyclic.
Then simple angle chasing gives $\triangle MED$ and $\triangle MKB$ similar.
So, $MK^2=ME.MB$
Similarly, $MK^2=MF.MC$
We are done.

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#4 h Aug 2, 2019, 6:10 AM • 3 Y

Y by top1csp2020, Adventure10, Mango247

Double BM to S, SG to J. Now BH/GS=AH/AG=MH/MG so B,S,M collinear so ABM=MSJ=EAJ so MD^2=ME.MB similar for F hence BEFC cyclic

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A-Thought-Of-God

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A-Thought-Of-God

#5 h Nov 8, 2020, 2:49 AM • 3 Y

Y by amar_04, Cindy.tw, Mango247

Let $\omega_1 \cap \omega_2 \cap \overline{BC}=K$ . Redefine $E$ as $E^*$ which is Humpty-Point WRT $\angle B$ in $\triangle ABD$ . Let $H_B$ be the orthocenter of $\triangle ABD$ . Let $X_B$ be the perpendicular foot from $B$ on $\overline{AD}$ . Now, notice that
$\begin{align*} \angle AE^*K+\angle ACK &= \angle AE^*H_B+\angle H_BE^*K+\angle ACB=180^\circ-\angle ADH_B+\angle H_BBK+\angle C \\ &=90+\frac{\angle A}{2}+\angle X_BBD+\angle C=90+\frac{\angle A}{2}+90-\frac{\angle A}{2}-\angle C+\angle C=180^\circ \end{align*}$
which gives that $\angle AE^*C=\angle AKC=90^\circ$ . So $E \equiv E^*$ . Analogously, we have $F$ Humpty-Point of $\triangle ACD$ WRT $\angle C$ . It's clear that $\angle MAE=\angle MBA$ which gives $MA^2=ME\cdot MB$ . Similarly $MA^2=MF\cdot MC$ . So $ME\cdot MB=MF\cdot MC,$ which finishes the problem. $\blacksquare$

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1896 posts

#6 h Nov 19, 2020, 4:56 AM • 1 Y

Y by A-Thought-Of-God

See my solution on my Youtube channel in the link below.

https://www.youtube.com/watch?v=zokW7q-ocgc

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VJ22

#7 h Jul 23, 2022, 6:55 PM

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Let External Angle Bisector of $\angle BAC$ intersect $(AC)$ , $(AB)$ at $X$ and $Y$ . Let $BX$ intersect $AD$ at $R$ . As, $(AB,AC;AR,AX)=-1$ . Also, $CX//AD$ , Let $BA$ intersect $CX$ at $T$ . So, Projecting from $A$ $(AB,AC;AR,AX)$ on $CX$ $\implies (C,T;X,\infty)=-1$ . This means $BX$ bisects $AD$ .So, $R=M$ . $\angle MEA=\angle XEA=\angle ACX=\angle CAD=\angle BAD$ . From here we get , $\triangle MAE \sim \triangle MBA$ . $\implies MA^2=MD^2=ME.MB$ , similarily, $\triangle MAF \sim \triangle MCA$ , So, $MA^2=MC.MF$ . $\implies ME.MB=MF.MC$ . Thus, $BEFC$ is cyclic.

This post has been edited 3 times. Last edited by VJ22, Jul 24, 2022, 10:56 AM
Reason: ,.

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#8 h Aug 24, 2023, 3:23 PM • 1 Y

Y by HoripodoKrishno

Beautiful problem! I just love the result! Me being such dumb nub, I fakesolved this initially and was jumping around until I realized that it was a fakesolve. This is why I gave up on synthetic tools and started applying stewart on this cuz I no longer had any motivation to continue... :ninja:

:ninja:

:rotfl:

$[asy] /* Converted from GeoGebra by User:Azjps using Evan's magic cleaner https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py */ /* A few re-additions are done using bubu-asy.py. This adds the dps, xmin, linewidth, fontsize and directions. */ pair A = (-57.30375,53.83617); pair B = (-83.34998,-39.17961); pair C = (43.19882,-39.61374); pair H = (-57.62315,-39.26787); pair D = (-31.07334,-39.35895); pair L = (-70.32686,7.32827); pair K = (-7.05246,7.11121); pair P = (-20.20788,-77.96332); pair X = (-22.03030,7.16259); pair Y = (61.56501,6.87581); pair M = (-44.18855,7.23860); pair I = (-40.28433,-6.63285); pair E = (-69.06976,-22.25320); pair F = (-23.36015,-3.92844); import graph; size(12cm); pen dps = linewidth(0.5) + fontsize(13); defaultpen(dps); real xmin = -130, xmax = 85, ymin = -100, ymax = 85; draw(A--B, linewidth(0.5)); draw(B--C, linewidth(0.5)); draw(C--A, linewidth(0.5)); draw(circle((-19.96366,-6.77439), 71.18934), linewidth(0.5)); draw(circle(L, 48.29685), linewidth(0.5) + blue); draw(circle(K, 68.61788), linewidth(0.5) + blue); draw(circle(P, 74.10196), linewidth(0.5) + linetype("4 4") + red); draw(B--M, linewidth(0.5) + red); draw(M--C, linewidth(0.5) + red); draw(A--P, linewidth(0.5)); draw(H--A, linewidth(0.5)); draw(L--Y, linewidth(0.5)); dot("$A$", A, N); dot("$B$", B, SW); dot("$C$", C, SE); dot("$H$", H, dir(270)); dot("$D$", D, SE); dot("$L$", L, NW); dot("$K$", K, NE); dot("$P$", P, dir(270)); dot("$X$", X, NW); dot("$Y$", Y, NE); dot("$M$", M, NW); dot("$I$", I, NW); dot("$E$", E, NW); dot("$F$", F, NE); clip((xmin,ymin)--(xmax,ymin)--(xmax,ymax)--(xmin,ymax)--cycle); [/asy]$

Let $P$ be the midpoint of the minor arc $\widehat{BC}$ and $I$ be the incenter of $\triangle ABC$ . Also, $H$ be the foot of the $A$ -altitude. Let $X$ and $Y$ be the intersections of $A$ -midline with $\odot(AHB)$ and $\odot(AHC)$ respectively such that $X$ and $Y$ . Also, let $K$ and $L$ be the midpoints of $AB$ and $AC$ respectively. All the lengths used in this solution are directed.

Firstly, we prove the following main claim.

$\textbf{CLAIM: }$ The inversion $\mathbf{I}(\odot(M,MA))$ swaps $\odot(AHB)\leftrightarrow\odot(AHC)$ .

$\textbf{PROOF: }$ Firstly, apply a suitable scale such that $LK=1$ . Also let $x=AL$ and $y=KA$ . Now from the angle bisector condition, we get that $LM=\dfrac x{x+y}$ and $MK=\dfrac{y}{x+y}$ Now we multiply everything by a scale factor of $(x+y)$ . So we finally have $AL=x(x+y)$ , $KA=y(x+y)$ , $LM=x$ and $MK=y$ . Thus it suffices to prove that $MA^2=MX\cdot MY$ . We now have, $MX=LX-LM=LA-LM=x(x+y)-x=x(x+y-1)$ and $MY=MK+KY=y+y(x+y)=y(x+y+1)$ . So $MX\cdot MY=xy(x+y+1)(x+y-1)$ . Finally by applying Stewart's Theorem on $\triangle ALK$ , we get that $MA^2=xy(x+y+1)(x+y-1)$ and we are done. $\square$

Firstly, we have $PD\cdot PA=PB^2$ . So the inversion $\mathbf{I}(\odot(P,PB^2)$ swaps $D\leftrightarrow A$ . This gives us that $\left\{\odot(BIC),\odot(AD)\right\}$ are orthogonal. So the inversion $\mathbf{I}(\odot(AD))$ fixes $\odot(BIC)$ . Now redefine $E=\odot(BIC)\cap\odot(ACH)$ and $F=\odot(BIC)\cap\odot(ABH)$ . Now we have that the inversion $\mathbf{I}(\odot(M,MA))$ fixes $\odot(BIC)$ . It also swaps $\odot(ABH)\leftrightarrow\odot(ACH)$ . So the intersection $E=\odot(BIC)\cap\odot(ACH)\leftrightarrow\odot(BIC)\cap\odot(ABH)=B$ and similarly $F\leftrightarrow C$ . So we get that $\overline{M-E-B}$ and $\overline{M-F-C}$ are collinear and we are done.

This post has been edited 1 time. Last edited by kamatadu, Aug 24, 2023, 3:24 PM

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#9 h Yesterday at 9:22 AM

Y by

Another headsolve today lets go.

We will $\sqrt{bc}$ invert at $A$ in $\triangle ABC$ and basically solve this.

Quote:

Let $N$ be minor arc midpoint of $\widehat{BC}$ and let $M=2N-A$ . Let $A'$ be $A$ -antipode. If Ray $\overline{CA'}$ intersects $(AMB)$ at $E$ and ray $\overline{BA'}$ intersects $(ACM)$ at $F$ . Prove that $(BFEC)$ is cyclic.

Claim: $E$ is reflection of $B$ over $\overline{NA'}$ and likewise.
Proof: We will phantom point this define $E$ as above). Obviously $E \in (AMB)$ as $ABEM$ is isoceles trapezoid ( $\overline{NA'}$ is common $\perp$ bisector of $\overline{AM}$ and $\overline{BE}$ ).

Now to prove $\overline{CA'E}$ is a line, see that $\begin{align*} \measuredangle AA'E &=\measuredangle AA'N+\measuredangle NA'E=\measuredangle ABN+\measuredangle BA'N \\ &=\measuredangle ABN+\measuredangle BCN=\measuredangle ABN+\measuredangle NBC=\measuredangle ABC=\measuredangle AA'C \end{align*}$ And done. $\square$

To finish see that $\overline{NA'}$ is common $\perp$ bisector of $\overline{BE}$ and $\overline{CF}$ , $BECF$ is isoceles trapezoid and hence cyclic.

This post has been edited 1 time. Last edited by ihategeo_1969, Yesterday at 9:22 AM

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