Let $ABCD$ be a cyclic quadrilateral inscribed in a circle $\Gamma$ such that $AB=AD$. Let $E$ be a point on the segment $CD$ such that $BC=DE$. The line $AE$ intersect $\Gamma$ again at $F$. The chords $AC$ and $BF$ meet at $M$. Let $P$ be the symmetric point of $C$ about $M$. Prove that $PE$ and $BF$ are parallel.
Problem
Source: 2021CHKMO Q3
Tags: geometry, cyclic quadrilateral
MMGMMGMMG
05.12.2020 12:51
(Fakesolve?)
Join $DF$, suppose $CD\cap BF=Q$
Claim 1: It suffices to prove that $CQ=QE$
Proof of Claim 1: My midpoint theorem, the desired result follows.
Claim 2: $M,C,F,E$ are concyclic.
Proof of Claim 2: $\measuredangle MCE=\measuredangle ACD=\measuredangle AFD=\measuredangle BFA= \measuredangle MFE$
Claim 3: $BD\parallel ME$
Proof of Claim 3: $\measuredangle (ME,CD)=\measuredangle MEC=\measuredangle MFC=\measuredangle BFC=\measuredangle BDC=\measuredangle (BD,CD)$
Let $R$ be the point on $DF$ such that $E,Q,F,R$ are concylic. Such point is unique.
Claim 4: $CQ=ER$
Proof of Claim 4: Trivially, $\Delta BCQ\cong \Delta DER$, the result follows.
Claim 5: $EQ=ER$
Proof of Claim 5: $AB=AD$, $\measuredangle RFE=\measuredangle DFA=\measuredangle AFB=\measuredangle EFQ$. The result follows.
Combining Claim 1, 4, and 5, we get the desired result.
Someone pointed out that Claim 2 and 3 are useless in my solution
But i wrote it during the test
and no it is not fakesolve
iman007
05.12.2020 15:58
ok nice problem i am going to post a trig bash solution.
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[/asy][/asy]
$\textbf{Claim 1:}$$CMEF \thickspace$ is cyclic.
$\textbf{Proof:}$
\[\angle{ABD}=\angle{ACD}=\angle{MCE}=\angle{BFA}=\angle{MFE}\]$\textbf{Claim 2:}$$\frac{BF}{FC}=\frac{Sin(\angle{CFK})}{Sin(\angle{KFE})}$
$\textbf{Proof:}$ By sines theorem we have
\[\frac{Sine(\angle{EFD})}{DE}=\frac{Sine(\angle{EDF})}{EF} \qquad (1)\]\[\frac{Sine(\angle{BFC})}{BC}=\frac{Sine(\angle{CBF})}{CF} \qquad (2)\]dividing $(1),(2)$,and using the intial conditions$BC=DE,\angle{BFA}=\angle{DFA}$,we get the desired result.
now by $\textbf{Claim 2}$ and using the "Ratio lemma" we get that $FB$ bisects $EC$.
$\boxed{\mathcal{Q,E,D}}$
iman007
05.12.2020 15:59
sorry for multi post but what is CHKMO??
k12byda5h
05.12.2020 16:02
Let $X = BF \cap AD$. It suffices that $X$ is the midpoint of $AE$ equivalent to $(A,C;B,F') = -1$ ($FF' \parallel CD$ and $F'$ lies on $\odot(ABC)$). Let $C'$ be a point on $CD$ such that $CA = C'A$. Then $\triangle ACB \cong \triangle AC'D$. Hence, $D$ is the midpoint of $C'E$. So, $-1=(E,C';D,\infty_{CD}) \overset{A}{=} (F,B'';D,A'') \overset{\text{reflect}}{=} (F',B;C,A)$. ($A'',B''$ are the reflection of $A,B$ across the perpendicular bisector of $CD$. Which can do angle chaseing that $B'',A,C'$ are collinear.)
Owleee
05.12.2020 16:07
iman007 wrote: sorry for multi post but what is CHKMO?? Hong Kong (China) Mathematical Olympiad
k12byda5h
05.12.2020 16:22
MMGMMGMMG wrote: (Fakesolve?)
Join $DF$, suppose $CD\cap BF=Q$
Claim 1: It suffices to prove that $CQ=QE$
Proof of Claim 1: My midpoint theorem, the desired result follows.
Claim 2: $M,C,F,E$ are concyclic.
Proof of Claim 2: $\measuredangle MCE=\measuredangle ACD=\measuredangle AFD=\measuredangle BFA= \measuredangle MFE$
Claim 3: $BD\parallel ME$
Proof of Claim 3: $\measuredangle (ME,CD)=\measuredangle MEC=\measuredangle MFC=\measuredangle BFC=\measuredangle BDC=\measuredangle (BD,CD)$
Let $R$ be the point on $DF$ such that $E,Q,F,R$ are concyclic. Such point is unique.
Claim 4: $CQ=ER$
Proof of Claim 4: Trivially, $\Delta BCQ\cong \Delta DER$, the result follows.
Claim 5: $EQ=ER$
Proof of Claim 5: $AB=AD$, $\measuredangle RFE=\measuredangle DFA=\measuredangle AFB=\measuredangle EFQ$. The result follows.
Combining Claim 1, 4, and 5, we get the desired result.
very nice solution
rafaello
05.12.2020 16:24
Since I have different trig bash from iman007, I post it. Claim. $CFEM$ is cyclic. $$\measuredangle MCE=\measuredangle ACD=\measuredangle BCA=\measuredangle BFA=\measuredangle MFE\qquad \square$$Let $X=BF\cap CD$. I show that $X$ is indeed the midpoint of $CE$. By Law of Sines,$$\frac{AE}{\sin{\angle ADE}}=\frac{DE}{\sin{\angle DAE}}$$and$$\frac{AC}{\sin{\angle ABC}}=\frac{BC}{\sin{\angle CAB}}.$$Dividing last two, we have that$$\frac{AE}{AC}=\frac{\sin{\angle CAB}}{\sin{\angle DAE}}=\frac{\sin{\angle CFM}}{\sin{\angle EMF}}=\frac{XM}{XE}.$$On the other hand,$$\frac{AE}{AC}=\frac{\sin{\angle ACE}}{\sin{\angle AEC}}=\frac{\sin{\angle MCE}}{\sin{\angle FMC}}=\frac{XM}{XC},$$hence $XC=XE$ and we are done.
iman007
05.12.2020 16:42
Owleee wrote: iman007 wrote: sorry for multi post but what is CHKMO?? Hong Kong (China) Mathematical Olympiad ohk nice!
hyay
05.12.2020 17:25
Let $Q \neq C$ be a point on $BC$ such that $BC = BQ$. Notice that $\triangle ADE \cong \triangle ABQ$. Then, $AE = AQ$ and $\angle EAQ = \angle DAB$, which implies $\angle AEQ = \angle ADB = \angle AFB$. So, $EQ \parallel BF$. Now let $P' = EQ \cap AC$. Since $P'Q \parallel BM$ and $BC = BQ$, we have $CM = MP'$. But $CM = MP$ aswell, so $P = P'$ and the conclusion $PE \parallel BF$ follows.
iman007
07.12.2020 18:13
iman007 wrote: ok nice problem i am going to post a trig bash solution.
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[/asy][/asy]
$\textbf{Claim 1:}$$CMEF \thickspace$ is cyclic.
$\textbf{Proof:}$
\[\angle{ABD}=\angle{ACD}=\angle{MCE}=\angle{BFA}=\angle{MFE}\]$\textbf{Claim 2:}$$\frac{BF}{FC}=\frac{Sin(\angle{CFK})}{Sin(\angle{KFE})}$
$\textbf{Proof:}$ By sines theorem we have
\[\frac{Sine(\angle{EFD})}{DE}=\frac{Sine(\angle{EDF})}{EF} \qquad (1)\]\[\frac{Sine(\angle{BFC})}{BC}=\frac{Sine(\angle{CBF})}{CF} \qquad (2)\]dividing $(1),(2)$,and using the intial conditions$BC=DE,\angle{BFA}=\angle{DFA}$,we get the desired result.
now by $\textbf{Claim 2}$ and using the "Ratio lemma" we get that $FB$ bisects $EC$.
$\boxed{\mathcal{Q,E,D}}$
i think my first claim was useless and i didn't really need that!
dgreenb801
09.12.2020 08:47
See my solution on my Youtube channel here: https://www.youtube.com/watch?v=NG8KEzYdb4Q]
Marinchoo
09.12.2020 17:06
We notice that ABCD is cyclic so $\angle ABC + \angle ADC = 180^{\circ}$. Now we rotate $\triangle ADE$ through point $A$ so that $D$ goes to $B$(we know this is possible as $AB=AD$). Let the point $Q$ be that point $E$ after the rotation, then $Q\in BC$ and $BQ=ED=BC$. Now $\angle MFE= \angle BFA = \angle ACD= \angle MAE \implies MCFE$ is cyclic and so $\angle FAD= \angle FCD= \angle FBD= \angle FME= \angle QAB$. Notice that $\angle MEC= \angle MFC= \angle BFC= \angle BAC$ and $\angle BCA= \angle ACD= \angle MCE$ and so $\triangle ABC \sim \triangle EMC$. Because $BC=BQ$ and $MC=MP$ so $\triangle EMP \sim \triangle ABQ$. Finally $\angle MEP= \angle BAQ=\angle EAD=\angle FCD=\angle FCE=\angle FME$ so $BF \| PE$.
Tintarn
16.12.2020 13:24
First of all, as has been mentioned before, we can kick out the points $M$ and $P$ by just noting that the claim is equivalent to $BF$ bisecting $CE$. So let $S$ be the intersection of $BF$ and $CE$.
We happen to not like the point $E$. How could we do better? Let's define $D'$ to be the point on $BC$ (outside the segment) such that $BC=CD'$.
Now $BCD'$ is isosceles and an easy angle chase shows that
\[\angle EFB=\angle AFB=\angle ADB=90^\circ-\frac{\angle BAD}{2}=\frac{\angle DCB}{2}=\angle CD'B=\angle ED'B\]so that $BEFD'$ is cyclic (note that we did not use the definition of $E$ so far).
But now of course
\[SC \cdot SD=SB \cdot SF=SE \cdot SD'\]and using that $SC+SD=SE+SD'$, this immediately shows that $SC=SE$ as desired.
KST2003
25.04.2021 18:50
Let $BF$ intersect $CD$ at $N$. Then it suffices to show that $N$ is the midpoint of $EC$. As $AD=AB$, $FA$ bisects $\angle DFB$, so by the angle bisector theorem, we have \[\frac{NE}{DE}=\frac{NF}{DF}=\frac{NC}{CB}\]and as $DE=BC$, we are done.
Sg48
24.07.2022 19:30
1. CMEF is cyclic => AM.AC = AE.AF 2. Triangle ACE is similar to Triangle BCM => CM/BC = CE/CA 3. Use BC = ED, CE.ED = AE.EF, CM = PM to obtain AM.AC = AE*AE 4. <AEP = <ACE = <AFM Hence Proved
lahmacun
25.07.2022 00:03
Let $\angle CBF=\angle CAF=\angle CDF=\alpha$ It suffices to show that $FB$ bisects $CE$, equivalently $$\frac{\sin{\frac{C}{2}}}{\sin{D}}=\frac{FC}{FE}$$Law of sines on triangles $CFB$ and $EFD$ respetively gives $$CF= CB\cdot\frac{\sin{\alpha}}{\sin{D}}$$and $$EF=ED\cdot \frac{\sin{\alpha}}{ \sin{\frac{C}{2}} }$$Dividing gives $$\frac{FC}{FE}=\frac{CB\cdot\frac{\sin{\alpha}}{\sin{D}}}{ED\cdot\frac{\sin{\alpha}} {\sin{\frac{C}{2}} }}=\frac{\sin{\frac{C}{2}}}{\sin{D}}$$
MathLuis
26.07.2022 05:09
Let $X$ a point in $\Gamma$ such that $BC=DE=DX$ (so by arcs it holds that $CX \parallel DB$) and let $EX \cap \Gamma=G$. Note that here we will define $P$ as the point in $CM$ such that $EP \parallel FB$, so we at last define $FM \cap CE=J$ and $N$ as midpoint or arc $CX$ in $\Gamma$. Claim 1: $ME \parallel BD$ Proof: Arc $BA$ is equal to arc $AD$ in $\Gamma$ so $\angle MCE=\angle MFE$ which gives $MCFE$ cyclic and Reims finishes. Claim 2: $XE$ is tangent to $(CFEM)$ Proof: Clearly $AN$ is diameter in $\Gamma$ so $\angle ADN=90$ which means that $\angle CEX=\angle XDA=\angle ABC=\angle CME$ so our claim is proven. Claim 3: Arcs $CG, BA, AD$ are equal in $\Gamma$ Proof: By Reims $(GEMA)$ is cyclic, also note that by Claim 2 we get $\angle CEG=\angle CFE$ which gives us that the arcs $XD, GA, BC$ are equal in $\Gamma$ so we get $CGBA$ ISLsusceles trapezoid so arcs $CG, BA, AD$ are equal as desired. Finishing: Becuase of the equality of the arcs in $\Gamma$ we get $\triangle CGE$ Isusceles so $GC,GE$ are tangent to $(MCFE)$ and since arcs $CG, BA$ are equal we get $FG, FJ$ isogonal w.r.t. $\angle CFE$ so $CJ=JE$ and by the parallels $CM=MP$ thus we are done