Show that AB/AC=BF/FC

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202 posts

#1 h Apr 2, 2012, 3:06 PM • 9 Y

Y by Davi-8191, nguyendangkhoa17112003, FaThEr-SqUiRrEl, centslordm, jhu08, megarnie, Adventure10, Mango247, and 1 other user

Let $ABC$ be an acute triangle. Denote by $D$ the foot of the perpendicular line drawn from the point $A$ to the side $BC$ , by $M$ the midpoint of $BC$ , and by $H$ the orthocenter of $ABC$ . Let $E$ be the point of intersection of the circumcircle $\Gamma$ of the triangle $ABC$ and the half line $MH$ , and $F$ be the point of intersection (other than $E$ ) of the line $ED$ and the circle $\Gamma$ . Prove that $\tfrac{BF}{CF} = \tfrac{AB}{AC}$ must hold.

(Here we denote $XY$ the length of the line segment $XY$ .)

This post has been edited 5 times. Last edited by syk0526, Apr 4, 2012, 6:48 AM

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Goutham

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Goutham

#2 h Apr 2, 2012, 4:11 PM • 14 Y

Y by wu2481632, IDMasterz, FaThEr-SqUiRrEl, centslordm, Wizard0001, BVKRB-, jhu08, CyclicISLscelesTrapezoid, megarnie, myh2910, jrsbr, Adventure10, Mango247, ehuseyinyigit

$\angle DAM=\angle DEM=\angle FAA'$ where $AA'$ is diameter of cirumcircle. So, $AF$ is symmedian...

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hatchguy

555 posts

hatchguy

#3 h Apr 2, 2012, 8:59 PM • 6 Y

Y by FaThEr-SqUiRrEl, centslordm, jhu08, megarnie, Adventure10, Mango247

Note that $\frac{BF}{FC} = \frac{\sin \angle BCF}{\sin \angle CBF} = \frac{\sin \angle BED}{\sin \angle CED} = \frac{BD \cdot CE}{CD \cdot BE}$

Clearly we have $\frac{BD}{DC} = \frac{ c \cos B}{b \cos C}$ .

On the other side, if $E'$ is the second intersection of $HM$ with the circumcircle of $ABC$ by an easy law of sines we can get that cuadrilateral $BECE'$ is harmonic and therefore $\frac{CE}{BE} = \frac {BE'}{CE} = \frac{AE \cdot cos C}{ AE \cdot cos B}$ and therefore we get $\frac{BF}{FC} = \frac{ c \cos B}{b \cos C} \cdot \frac{AE \cdot cos C}{ AE \cdot cos B} = \frac{AB}{AC}$ and we are done.

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kakkokari

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kakkokari

#4 h Apr 2, 2012, 9:03 PM • 13 Y

Y by nguyendangkhoa17112003, AlastorMoody, FaThEr-SqUiRrEl, centslordm, jhu08, CyclicISLscelesTrapezoid, Adventure10, Mango247, ehuseyinyigit, and 4 other users

//cdn.artofproblemsolving.com/images/8b1ad0c683d84680f3f50c3663dc4546119dcab0.png

Well, this question is like the picture shows and i used butterfly theorem during the test without proving it...

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WakeUp

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WakeUp

#5 h Apr 2, 2012, 9:27 PM • 7 Y

Y by pmathgol, FaThEr-SqUiRrEl, centslordm, jhu08, Adventure10, Mango247, and 1 other user

Extend $FC$ beyond $C$ such that $CF=CG$ . This is one of many threads that prove the well-known result that if $A'$ is the antipode of $A$ on $(ABC)$ then $H,M,A'$ are collinear. Therefore $\angle AEM=\angle AEA'=90^{\circ}=\angle ADM$ so $AEDM$ is cyclic. Therefore $\angle AMB=\angle AED=\angle AEF=\angle ACF$ . Also clearly $\angle ABM=\angle AFC$ . It follows that $\triangle ABM\sim AFC$ . It follows by SAS that $\triangle AMC\sim\triangle ACG$ . Therefore $\angle ACM=\angle AGC$ which implies $\angle AFB=\angle AGC$ . Also $\angle ABF=\angle ACG$ so we have $\triangle ABF\sim\triangle ACG$ . So $\frac{AB}{BF}=\frac{AC}{CG}=\frac{AC}{CF}$ and the results follows.

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cwein3

148 posts

cwein3

#6 h Apr 3, 2012, 6:58 AM • 4 Y

Y by FaThEr-SqUiRrEl, centslordm, jhu08, Adventure10

Extend $EH$ to intersect $\Gamma$ again $Q$ . Then it's well known that $AQ$ is a diameter, so $\angle AEQ = 90$ . Thus, $AEDM$ is cyclic, so $\angle MAD = \angle MED = \angle QBF$ . Furthermore, $\angle QBC = \angle QAC = \angle DAB$ . Thus, $\angle CBF = \angle QBC + \angle QBF = \angle MAD + \angle DAB = \angle MAB$ , so $AF$ is the symmedian and the conclusion easily follows.

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Ali-mes

37 posts

Ali-mes

#7 h Apr 6, 2012, 9:37 PM • 4 Y

Y by FaThEr-SqUiRrEl, centslordm, jhu08, Adventure10

Here is my ugly non-finished solution

:

First we will show that: $\angle{AEM}=\frac{\pi}{2}$ . Consider point $X$ the second intersection of line $(AD)$ with $(ABC)$ , and point $Y$ such that: $\overrightarrow{BY}=\overrightarrow{HC}$ .
Clearly, $BHCY$ is a parallelogram, so: $\angle{BYC}=\angle{BHC}=\pi-\angle{A}$ , therefore: $Y \in (ABC)$ .
And, it's a very well-known fact that $D$ is the midpoint of $[HX]$ , consequently, by the converse of Thales' theorem in $\triangle HXY$ : $(MD) \parallel (XY)$ .
Thus: $\angle{AEM}=\angle{AEY}=\angle{AXY}=\angle{HXY}=\angle{HDM}=\angle{ADC}=\frac{\pi}{2}$ .

And since $\angle{ADM}=\frac{\pi}{2}=\angle{AEM}$ , the quadrilateral $AEDM$ is cyclic.
Now, by simple angle chasing, we get: $\angle{BCF}=\angle{BEF}=\angle{AEB}-\angle{AED}=\pi-\angle{C}-\pi+\angle{AMB}=\angle{AMB}-\angle{C}$ .
And: $\angle{CBF}=\angle{CEF}=\angle{CED}=\angle{AED}-\angle{AEC}=\pi-\angle{AMB}-\angle{B}$ .

Therefore: $\frac{BF}{FC}=\frac{\sin(\angle{BCF})}{\sin(\angle{FBC})}=\frac{\sin(\angle{AMB}-\angle{C})}{\sin(\angle{AMB}+\angle{B})}$ .

So it's enough to show: $\frac{\sin(\angle{AMB}-\angle{C})}{\sin(\angle{AMB}+\angle{B})}=\frac{\sin(\angle{C})}{\sin(\angle{B})}$ And this can be proved using metric relations...

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dinoboy

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dinoboy

#8 h Apr 6, 2012, 10:19 PM • 5 Y

Y by FaThEr-SqUiRrEl, centslordm, jhu08, Adventure10, and 1 other user

Sorry if my proof is basically equivalent to some above proof. Note that whenever I say similarity suffices, I mean similarity in the way which would imply the problem (and I do prove that similarity each time).

Define $A'$ to be the point where $OA$ hits $\Gamma$ that is not $A$ . I claim $H,M,A'$ are collinear. To prove this, we use vectors. Denote $O$ as the origin and we have $M = \frac{B + C}{2}$ and $H = A + B + C$ by well-known results. Then remark that $2M + (1-2)H = -A = A'$ , therefore $H,M,A'$ are collinear and we are done with this proof.
Now, remark that $\angle AEM = \angle AEA' = 90$ because that $AA'$ is a diameter. Then as $\angle ADM = 90$ , we have $AEDM$ is cyclic.
Rearrange what we want to show as $BF/AB = CF/AC$ . Clearly by the diagram $AFB \not \sim AFC$ so that approach would be hopeless. However, remark that if we extend $CF$ to a point $X$ such that $CX = CF$ , then the condition is $BF/AB = CX/AC$ and then since $ABFC$ is cyclic we have $\angle ACX = \angle ABF$ , therefore the problem statement holds iff $\triangle ACX \sim \triangle ABF$ .

To show this, remark that $\angle BFA = \angle BCA = \angle MCA$ and that $\angle AMC = 180 - \angle AMD = \angle AEF = \angle ABF$ in the case of $D$ lies on the left of $M$ . The case of $D$ lies on the right we have $\angle AMC = \angle AMD = 180 - \angle AED = 180 - \angle AEF = \angle ABF$ . In either case we find $\triangle BFA \sim \triangle AMC$ . I claim that $\triangle AMC \sim \triangle ACX$ .

To prove this assertion, first remark that $MC/CX = MC/CF$ . If we can show $MC/CF = AM/AC$ then we would be done because $\angle ACG = \angle AMC$ . Re-arranging we have and noting $MC = MB$ we have $MB/AC = AM/CF$ . Then remark that $\angle AMB = \angle AMD = 180 - \angle AED = 180 - \angle AEF = \angle ACF$ when $D$ is on the left of $M$ and $\angle AMB = 180 - \angle AMD = \angle AED = \angle AEF = \angle ACF$ so it suffices to show $\triangle ACF \sim \triangle AMB$ . To show this we only need to show one more angle relation. But $\angle ABM = \angle ABC = \angle AFC$ and therefore we are done.

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javlik

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javlik

#9 h Apr 7, 2012, 7:39 AM • 5 Y

Y by FaThEr-SqUiRrEl, centslordm, jhu08, Adventure10, Mango247

Nice problem!

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sunken rock

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sunken rock

#10 h Apr 7, 2012, 8:06 PM • 6 Y

Y by FaThEr-SqUiRrEl, centslordm, jhu08, Adventure10, Mango247, and 1 other user

If $AA'$ is a diameter of the circle $\odot (ABC)$ , then $BA'CH$ is a parallelogram, $E, H, M, A'$ collinear, hence $\angle AEA'=90^\circ$ , or $AEDM$ cyclic, with $\angle ADE=\angle AME\ (\ *\ )$ . If $AD$ intersects circle $(ABC)$ at $Q$ , from $\angle ABD=\angle CAA'$ , we get $Q$ the reflection of $A'$ w.r.t. midpoint of the arc $BC$ .
If $AM$ intersects the circle $(ABC)$ at $P$ , from $(*)$ we get that the arcs $PA', QF$ are congruent, hence $P$ and $F$ are symmetrical about the midpoint of the arc $BC$ , i.e. $AF$ is the symmedian, done.

Best regards,
sunken rock

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iban001

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iban001

#11 h Apr 13, 2012, 10:34 AM • 5 Y

Y by FaThEr-SqUiRrEl, jhu08, Adventure10, Mango247, ehuseyinyigit

I prove it with a complex method which has inversion and butterfly theorem ....

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iban001

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iban001

#12 h Apr 13, 2012, 10:37 AM • 5 Y

Y by Konigsberg, FaThEr-SqUiRrEl, jhu08, Adventure10, Mango247

kakkokari wrote:

http://i1178.photobucket.com/albums/x361/cltcltxb/APMOP4.png

Well, this question is like the picture shows and i used butterfly theorem during the test without proving it...

you just need to prove AFOM is cyclic.your method is similar to mine.

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hyperspace.rulz

287 posts

hyperspace.rulz

#13 h Apr 20, 2012, 1:45 AM • 3 Y

Y by DakuMangalSingh, jhu08, Adventure10

You can also show that AEDM is cyclic using Power of a Point. As was proven earlier in this thread, HD extended meets the circumcircle again at a point X such that HD=DX and HM extended meets the circumcircle at a point Y such that HM=MY. But A, E, X and Y all lie on the circumcircle, so by Power of a Point:

$HA \cdot HX = HE \cdot HY$

$\Rightarrow HA \cdot 2HD = HE \cdot 2HM$

$\Rightarrow HA \cdot HD = HE \cdot HM$

Therefore, by the converse of Power of a Point, AEDM is cyclic. From here we can angle chase out the result - firstly a simple angle chase yields $\angle CAM=\angle BAF$ and hence AF is a symmedian. It follows that ABFC is harmonic and therefore $AB \cdot CF=AC \cdot BF$ , from which the result immediately follows.

Cheers,
Hyperspace Rulz!

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Invader_2011

53 posts

Invader_2011

#14 h Jun 26, 2012, 12:58 AM • 5 Y

Y by applepi2000, jhu08, Adventure10, Mango247, ehuseyinyigit

Proof.

Let line $AD$ intersect the circumcircle at $H'$ , and let line $HM$ intersect the circumcircle at $N$ .

Since $H'N$ is parallel to $BC$ , we have that the pencil $(NH', NM; NB, NC)$ is harmonic.

Intersecting this pencil at the circumcircle we get that the quadrilateral $EBH'C$ is harnomic.

then the pencil $D(E, H'; B, N)$ is harmonic.

Intersecting this pencil at the circumcircle we get that the quadrilateral ABFC is harmonic. So $\frac{AB}{AC}=\frac{BF}{CF}$ . Thus we are done.

Solution collaborated with applepi2000.

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vslmat

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vslmat

#15 h Aug 22, 2012, 3:26 PM • 4 Y

Y by jhu08, Adventure10, Mango247, and 1 other user

Let $O$ be the circumcenter of $ABC, AO$ cut the circumcircle again at $G$ .
As $GC\perp AC, GC\parallel BH$ and it is easy to see well known facts that $GC = BH$ (see the diagram) and $\angle HBC = \angle BCG = 90^{\circ} - \angle ACB$ , moreover, $BM=CM$ , hence $H, M, G$ are collinear.
Notice that $\angle FCG = \angle FEG = \angle DAM$ , hence $\angle AMB =\angle ACF$
$\Delta ABM\sim \Delta AFC$ , thus $\frac{AB}{AF} = \frac{BM}{FC}$ (*)
Since $\angle ABF =\angle AMC = 180^{\circ}-\angle ACF$ (or $\angle AMB$ ), and $\angle ABC =\angle AFC,\Delta ABF\sim \Delta AMC$ , thus $\frac{AC}{AF} = \frac{CM}{BF}$ (**)
As $BM = CM$ , (*) and (**) give $\frac{AB}{AC} =\frac{BF}{CF}$

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ZNatox

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ZNatox

#65 h Mar 4, 2023, 4:44 PM

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Literally one application of Pascal's theorem. Let $K=(AM) \cap (ABC), G=(AD) \cap (ABC), A'=(MH) \cap (ABC) \neq E$ . We apply Pascal's theorem on $AGA'EFK$ . Then $(AG) \cap (EF)=D, (GA')\cap (FK), (A'E)\cap(KA)=M$ all lie on a line. But $(AG)$ and $(AA')$ are isogonal, which implies $(GA')\parallel (BC) \equiv (DM)$ . Hence, $(FK)$ is parallel to $(BC)$ , which implies the desired equality.

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Knty2006

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Knty2006

#67 h Apr 19, 2023, 3:03 AM

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Note $\angle{AEM}= \angle{ADM}=90$
So, we have $A,E,D,M$ cyclic
$\angle{ABF}=\angle{AEF}=90+\angle{FEM}=90+\angle{DEM}=90+\angle{DAM}=\angle{AMC}$

So, $\triangle{ABF}=\triangle{AMC}$ which implies $F$ is the $A$ symmedian.

Q.E.D

This post has been edited 1 time. Last edited by Knty2006, Apr 19, 2023, 3:04 AM
Reason: Oops

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YaoAOPS

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YaoAOPS

#68 h Jun 12, 2023, 9:02 PM

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Note that the result is equivalent to $AF$ being the $A$ -Symmedian.
We claim that quadrilateral $MDEA$ is cyclic. This follows since $HM \cdot HE = HD \cdot HA$ follows from reflecting the orthocenter.
Then, $\measuredangle MAC = \measuredangle MAE + \measuredangle EAC = \measuredangle MDE + \measuredangle EFC = \measuredangle CDF + \measuredangle DFC = \measuredangle DCF = \measuredangle BAF$ gives the result.

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asdf334

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asdf334

#69 h Jul 11, 2023, 12:16 AM

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Project through $D$ , then through $M$ ... resulting quadrilateral is obviously harmonic

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john0512

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john0512

#70 h Aug 16, 2023, 6:20 AM

Y by

It is well known that $E$ is the A-Queue point $(ABC)$ intersect $(AEF)$ . Furthermore, by radical axis theorem on $(AEF),(ABC),(BCEF)$ , we have that $AE,FE,BC$ concur at a point, say $P$ . Thus, we have $(AF;BC)^E=(PD;BC)=-1$ by projecting from $E$ onto line $BC$ , hence done.

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shendrew7

794 posts

shendrew7

#71 h Aug 27, 2023, 3:37 AM

Y by

Due to monotonicity, there exists a unique point $F'$ on minor arc $AB$ such that $\frac{BF'}{CF'} = \frac{AB}{AC}$ , which is well known to be the intersection of the $A$ -symmedian with $(ABC)$ . We want to show $F$ is this point, or $\angle BAF = \angle MAC.$
Denote $A'$ as the antipode of $A$ on $(ABC)$ . Then it is well known that $H$ , $M$ , and $A'$ are collinear, so $\angle AEM = \angle AEA' = 90$ .

Claim 1: $\angle BAD = \angle A'AC$ .

We may simply state $H$ and $O$ are isogonal conjugates, or
$\angle BAD = 90 - \angle ABC = 90 - \angle AA'C = \angle A'AC.~\Box$
Claim 2: $\angle ADE = \angle AME$ and $\angle AFE = \angle AA'E$ .

This follows directly from cyclic quadrilaterals $DEAM$ and $AEFA'$ . $\Box$

From these results, we find

$\begin{align*} \angle BAF &= \angle BAD + \angle DAF \\ &= \angle A'AC + \left(\angle ADE - \angle AFE\right) \\ &= \angle A'AC + \left(\angle AME - \angle AA'E\right) \\ &= \angle A'AC + \angle MAA' \\ &= \angle MAC.~\blacksquare \end{align*}$
$[asy] size(225); defaultpen(linewidth(0.4)+fontsize(10)); pair A, B, C, H, D, M, E, F, A1; A = dir(120); B = dir(210); C = dir(330); H = orthocenter(A, B, C); D = foot(A, B, C); M = .5B + .5C; E = IP(circumcircle(A, B, C), H--5H-4M); F = IP(circumcircle(A, B, C), D--3D-2E); A1 = -A; draw(E--A--B--C--A--F--E--A1--A--D--A--M); draw(circumcircle(A, B, C)); markscalefactor = .02; filldraw(anglemark(B, A, D)^^anglemark(A1, A, C), cyan); markscalefactor = .015; filldraw(anglemark(A, D, E)^^anglemark(A, M, E), blue); filldraw(anglemark(A, F, E)^^anglemark(A, A1, E), green); markscalefactor = .008; draw(rightanglemark(A1, E, A)^^rightanglemark(C, D, A)); label("$A$", A, N); label("$B$", B, SW); label("$C$", C, SE); label("$H$", H, NE); label("$D$", D, S); label("$M$", M, S); label("$E$", E, W); label("$F$", F, S); label("$A'$", A1, SE); dot(A); dot(B); dot(C); dot(H); dot(D); dot(M); dot(E); dot(F); dot(A1); [/asy]$

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eibc

600 posts

eibc

#72 h Aug 28, 2023, 1:39 AM

Y by

$[asy] unitsize(4cm); pair A = dir(110); pair B = dir(205); pair C = dir(335); pair D=foot(A, B, C); pair BB=foot(B, C, A); pair CC=foot(C, A, B); pair H=A+B+C; pair M=(B+C)/2; pair G=IP(Line(M, H, 10),unitcircle,1); pair T=extension(A, G, B, C); pair F=IP(Line(D, G, 10), unitcircle, 0); pair I=IP(Line(D, A, 10), unitcircle, 0); draw(A--B--C--cycle, heavygreen); draw(A--T--BB, red); draw(T--C, red); draw(G--F, heavygreen); draw(A--I, heavygreen); draw(G--M, heavygreen); draw(circumcircle(A, G, H), red); draw(circumcircle(A, B, C), heavygreen); draw(circumcircle(B, CC, C), red); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$B'$", BB, NE); dot("$C'$", CC, W); dot("$H$", H, NE); dot("$M$", M, dir(M)); dot("$G$", G, dir(G)); dot("$T$", T, dir(T)); dot("$F$", F, SE); dot("$$", I, SW); [/asy]$

oops pretend $E$ was renamed to $G$

Let $B', C',$ be the feet of the altitudes in $\triangle ABC$ from $B$ and $C$ . Since $MH$ passes through the $A$ -antipode of $(ABC)$ we find that $G$ lies on $(AB'HC')$ . Also, by radical center on $(AB'HC'), (BCB'C')$ , and $(ABC)$ we find that $B'C'$ , $AG$ , and $BC$ concur at some point $T$ . Then by Ceva-Menelaus, we have
$-1 = (T, D; B, C) \overset{A}{=} (G, \overline{AD} \cap (ABC); B, C) \overset{D}{=} (F, A; C, B),$ which implies that $ABFC$ is harmonic.

This post has been edited 2 times. Last edited by eibc, Sep 18, 2023, 7:31 PM
Reason: added diagram

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Inconsistent

1455 posts

Inconsistent

#73 h Mar 4, 2024, 5:25 AM

Y by

Not a bad problem. Perhaps when it was first proposed, this config was still fresh. Nowadays, it is textbook.

Let $A'$ be the antipode of $A$ , then it follows from $\overline{HMA'}$ that $\angle AEA' = \angle AEH = 90^{\circ}$ , so let $V, W$ be the feet of the altitudes from $B, C$ in $\triangle ABC$ , then we have $(AEHVW)$ . Let $R = VW \cap BC$ , then by radical axis theorem it follows that $AE, VW, BC$ concur at $R$ . Now by Ceva-Menelaus theorem it follows that $-1 = (R, D; B, C) \stackrel{E}= (A, F; B, C)$ , so we are done.

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Shreyasharma

678 posts

Shreyasharma

#74 h Mar 4, 2024, 6:19 AM

Y by

Ok.

Note that $E$ is the $A$ -queue point. Projecting through $E$ we find that,
$\begin{align*} (AF, BC) \overset{E}{=} (\overline{AE} \cap \overline{BC}, D; BC) = -1 \end{align*}$ Thus $ABFC$ is harmonic which implies the conclusion.

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Aiden-1089

278 posts

Aiden-1089

#75 h Mar 4, 2024, 3:09 PM

Y by

Let $A'$ and $H'_A$ denote the intersection between $(ABC)$ and rays $HM$ , $HD$ respectively. It is well-known that $BC \parallel A'H'_A$ . Since
$(B,C;A,F) \stackrel{D}{=} (C,B;H'_A,E) \stackrel{A'}{=} (C,B;\infty_{BC},M) = -1,$ $ABFC$ is a harmonic quadrilateral so we are done. $\square$

This post has been edited 1 time. Last edited by Aiden-1089, Mar 4, 2024, 3:10 PM
Reason: typo

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joshualiu315

2513 posts

joshualiu315

#76 h May 5, 2024, 1:24 AM

Y by

It is well-known that $H$ reflected over $M$ is the antipode of $A$ , call it $A'$ . Hence, $\angle MEA = \angle MDA = 90^\circ$ , so $AMDE$ is cyclic. This means that

$\angle MAD = \angle MED = \angle FAA'.$
It is well-known that $\overline{AD}$ and $\overline{AA'}$ are isogonal with respect to $\angle BAC$ . This means that $\angle CAA' = \angle BAD$ . Thus,

$\angle BAM = \angle BAD + \angle MAD = \angle CAA' + \angle FAA' = \angle CAF,$
which implies that $\overline{AM}$ and $\overline{AF}$ are isogonal with respect to $\angle BAC$ . This means $\overline{AF}$ is the $A$ -symmedian of $\triangle ABC$ , so $ABFC$ is a harmonic quadrilateral. Thus, we have $\tfrac{BF}{CF} = \tfrac{AB}{AC}$ , as desired. $\square$

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blueberryfaygo_55

340 posts

blueberryfaygo_55

#77 h May 14, 2024, 10:39 PM • 3 Y

Y by KevinYang2.71, Orthogonal., megarnie

Claim: $AEDM$ is cyclic.
Proof. We first present the following lemma:

Lemma: In $\Delta ABC$ , let $H$ be the orthocenter and $M$ be the midpoint of $BC$ . Let $X$ be a point on $HM$ such that $M$ is the midpoint of $HX$ . Then, $AX$ is a diameter in $(ABC)$ .
Proof. Let $Y$ be the reflection of $H$ across $BC$ , $D$ be the projection of $A$ on $BC$ , and $E$ be the projection of $B$ on $AC$ . Since $B$ lies on the perpendicular bisector of $HY$ , we have $\begin{align*} \angle BYH &= \angle BHY \\ &= \angle AHE \\ &= 90^{\circ} - \angle DAC \\ &= \angle ACB \end{align*}$ which implies that $ABYC$ is cyclic. However, we know that $MD$ is the $H$ -midline in $\Delta HYX$ , so $DM \parallel XY$ , implying that $\angle AYX = 90^{\circ}$ . Further, by construction, $BHCX$ is a parallelogram, so $BH \parallel CX$ , or $BE \parallel CX$ , implying that $\angle ACX = \angle AYX = 90^{\circ}$ , so the result follows. $\blacksquare$

Returning to the original problem and applying the lemma by letting $X$ be the reflection of $H$ across $M$ , we know that $AX$ is a diameter in $\Gamma$ , so $\angle AEX = 90^{\circ} = \angle ADM$ and the desired condition follows. $\blacksquare$

Then, we have $\begin{align*} \angle AMB &= 180^{\circ} - \angle AED \\ &= 180^{\circ} - \angle AEF \\ &= \angle ACF \end{align*}$ by angle chasing in cyclic quadrilaterals. However, we know that $\angle ABM = \angle ABC = \angle AFC$ , so it follows that $\Delta ABM \sim \Delta AFC$ by angle-angle similarity, implying that $\dfrac{AM}{MB} = \dfrac{AC}{CF}$ and $\angle BAM = \angle CAF$ , which in fact gives $\angle BAF = \angle CAM$ . Thus, since $\angle AFB = \angle ACB$ , we also obtain $\Delta ABF \sim \Delta AMC$ , giving $\begin{align*} \dfrac{AB}{BF} &= \dfrac{AM}{MC} \\ &= \dfrac{AM}{MB} \\ &= \dfrac{AC}{CF} \end{align*}$ which is also the desired conclusion. $\blacksquare$

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Eka01

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Eka01

#78 h Aug 27, 2024, 3:24 PM

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Let $X$ be the $A$ expoint of $\Delta ABC$ . Then it is well known that $E$ is the $A$ queue point and $X,E,A$ lie on a line and that $(XD;BC)=-1$ .
$\implies -1=(XD;BC) \stackrel{E}{=} (AF;BC)$ And we are done.

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cj13609517288

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cj13609517288

#79 h Feb 23, 2025, 7:55 PM

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Project the harmonic bundle $(B,C;D,AE\cap BC)$ through $E$ . $\blacksquare$

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AshAuktober

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AshAuktober

#80 h Apr 2, 2025, 2:04 PM

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Yay!
Note that $(B,C,A,F) \stackrel{E}{=} (B,C,AE\cap BC, D).$ Now $AE, BC$ meet at the harmonic conjugate of $D$ from Radax and Ceva-Menelaus, so we're done.

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