Arc Midpoints Form Cyclic Quadrilateral

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ike.chen

1162 posts

ike.chen

#1 h Jul 9, 2023, 4:36 AM • 5 Y

Y by bjump, Rounak_iitr, deplasmanyollari, lian_the_noob12, eggymath

In the acute-angled triangle $ABC$ , the point $F$ is the foot of the altitude from $A$ , and $P$ is a point on the segment $AF$ . The lines through $P$ parallel to $AC$ and $AB$ meet $BC$ at $D$ and $E$ , respectively. Points $X \ne A$ and $Y \ne A$ lie on the circles $ABD$ and $ACE$ , respectively, such that $DA = DX$ and $EA = EY$ .
Prove that $B, C, X,$ and $Y$ are concyclic.

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MarkBcc168

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MarkBcc168

#2 h Jul 9, 2023, 4:38 AM • 4 Y

Y by khina, StarLex1, Rounak_iitr, Aliiiiii1188

This is ISL 2022 G2.

Solution

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SQTHUSH

154 posts

SQTHUSH

#3 h Jul 9, 2023, 4:54 AM

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suppose $DP\cap \odot(ABD)=\{D,U\},EP\cap \odot(AEC)=\{E,V\},\odot(ABD)\cap \odot(ACE)=\{A,Z\}$
$\angle UAV=\angle UAB+\angle BAC+\angle CAV=180^{\circ}$
Which means that $U-V-A$
So $U,E,D,V$ are cyclic $\Rightarrow PD\cdot PU=PZ\cdot PV$
Hence $A-P-Z$
Let $AZ\cap XB=T$
Notice $\angle ABD=TBC$
Which means that points $A$ and $T$ are symmetric about $BC$
So $\angle ACB=\angle BCT$
Which means that $T-Y-C$
So $TB\cdot TX=TZ\cdot TA=TY\cdot TC$
Hence $X,B,Y,C$ are cyclic

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VicKmath7

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VicKmath7

#4 h Jul 9, 2023, 7:30 AM • 1 Y

Y by TheStrayCat

Solution of G2

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rstenetbg

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rstenetbg

#5 h Jul 9, 2023, 7:55 AM

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$WLOG$ let $AB>AC$ . Let $\omega_{1}=(ABD)$ and $\omega_{2}=(AEC)$ .

Claim 1: $AF\equiv \rho(\omega_{1}, \omega_{2})$
Proof: It is enough to show that $deg(F,\omega_{1}) = deg(F,\omega_{2})$ $\Leftrightarrow FD\cdot FB = FE\cdot FC$ $\Leftrightarrow \frac{FE}{FB} = \frac{FD}{FC}$ . However, $\frac{FE}{FB} =\frac{FP}{FA}$ because $EP||AB$ and $\frac{FD}{FC}=\frac{FP}{FA}$ because $DP||CA$ . Hence $\frac{FE}{FB} = \frac{FD}{FC} = \frac{FP}{FA}$ .

Now let $K$ be the symmetrical point of $A$ wrt to $BC$ $\Rightarrow K\in \rho(\omega_{1}, \omega_{2})$ .

Claim 2: $K\in BX$ and $K\in CY$

Proof: From $\angle DBA = \angle DXA = \angle DAX = \angle DBX = \beta$ and from $\angle KBF = \angle ABF = \beta$ , we get that $\angle KBF = \angle XBF$ . So $K\in BX$ . Similarly, we get that $K\in CY$ .

Now $BXCY$ is cyclic $\Leftrightarrow KX\cdot KB = KC\cdot KY \Leftrightarrow deg(K,\omega{1}) = deg(K,\omega{2})$ , but this is true since $K\in \rho(\omega_{1}, \omega_{2})$ .

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Assassino9931

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Assassino9931

#6 h Jul 9, 2023, 8:03 AM

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From Thales we have $\frac{FB}{FE} = \frac{FA}{FP} = \frac{FC}{FD}$ , so $FB \cdot FD = FE \cdot FC$ , i.e. $F$ lies on the radical axis of the circles of $ABD$ and $ACE$ . So if $T$ is the second common point of these circles, then $A$ , $F$ , $T$ are collinear.

Now note that if it turns out that $BX$ , $CY$ and $AFT$ are concurrent, then from Power of a Point $XBYC$ would be cyclic. To argue the latter concurrency, let $BX \cap AFT = U$ . With $\angle ABC = \beta$ we have $\angle ABX = \angle ADX = 180^{\circ} - 2\angle AXD = 180^{\circ} - 2\angle ABD = 180^{\circ} - 2\beta$ (relying on the given $DX = DA$ ), so $\angle UBF = 180^{\circ} - \angle ABC - \angle ABX = \beta = \angle ABC$ . Therefore $ABU$ is isosceles and $BC$ is the perpendicular bisector of $AU$ . Analogously if $CY \cap AFT = V$ , then $BC$ is the perpendicular bisector of $AV$ . Therefore $U\equiv V$ and we are done.

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demmy

133 posts

demmy

#7 h Jul 9, 2023, 8:28 AM • 4 Y

Y by soelinhtetptn20204, SatisfiedMagma, Rounak_iitr, Mathematician2080

Nice problem, although it is trivial using geogebra

Claim : $(ABD) \cap (ACE)$ lies on lines $AP$

Proof : Let $K = AP \cap (ABD)$ and $J$ is foot from $A$ to $BC$
it suffices to prove that $K \in (ACE)$
By power of point ; $AJ \times JK = BJ \times JD$
Since $\frac{AJ}{JP} = \frac{BJ}{JE}$
Divided first equation by second equation :
$JP \times JK= JE \times JD$ $\rightarrow K,E,P,D$ concylic
then $\angle EKJ = \angle JDP = \angle JCA$
Thus $E$ lies on $(ACE)$ $\blacksquare$

Let $I = BX \cap CY$

By some angle chasing implies that $\triangle BIC \equiv \triangle BAC$
Then $I \in AP$
Which means $A-P-K-I$

By Claim implies $(ABD) \cap (ACE) \in AP$
Power of point : $IB \times IX = IK \times IA = IY \times IC$
Hence $B,X,C,Y$ concyclic. $\blacksquare$

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NO_SQUARES

1074 posts

NO_SQUARES

#8 h Jul 9, 2023, 9:05 AM

Y by

Nice problem!
Let $XB \cap YC = T$ . By angle chasing easy to show that $A, P, T$ are collinear. So, by PoP we need to prove that $A,K,P$ are collinear, where $K$ is second intersection of $(ABD),(ACE)$ or (by PoP) that $BH \cdot HD = CH \cdot HE$ where $H = AP \cap BC$ , which is obvious because by Thales $\frac{HD}{HC}=\frac{HP}{HA}=\frac{HE}{HB}$

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g0USinsane777

44 posts

g0USinsane777

#9 h Jul 9, 2023, 9:12 AM

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(1) Let F be the foot of perpendicular from $A$ to $BC$ . Similarity yields $\frac{FC}{FD}=\frac{FA}{FP}=\frac{FB}{FE} => FB\times FD=FC\times FE$
(2) Let $G$ be the second intersection of $(ABD)$ and $(ACE)$ and let $F'=AG\cap BC$ . By POP, $F'B \times F'D = F'C \times F'E$
From (1) and (2), we get $F'=F$ . Hence, $F$ lies on the radical axis $AT$ of $(ABD)$ and $(ACE)$ .
Let $BX \cap CY=A'$
By simple angle chasing we get that $\angle ABC = \angle A'BC$ and $\angle ACB = \angle A'CB=>A'$ is the reflection of $A$ w.r.t. $BC$ and $A'$ lies on radical axis $AG$ which gives us $A'B \times A'X = A'Y \times A'C$ .
Hence, $B,X,C,Y$ are concyclic.

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pokmui9909

185 posts

pokmui9909

#10 h Jul 9, 2023, 1:40 PM

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$AF \cap (ABD) = K \implies AF \cdot FK = BF \cdot FD = EF \cdot FC \implies A, E, K, C$ concyclic.
Let $T (\neq A)$ be a point on $AK$ such that $AF = FT$ .
$DA = DX \implies \angle ABC = \angle XBC$ , so $B, X, T$ are collinear. Similarly, $C, Y, T$ are collinear.
$\therefore BX, AK, CY$ are concurrent at $T \implies TX \cdot TB = TK \cdot TA = TY \cdot TC \implies B, X, Y, C$ are concyclic.

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popop614

270 posts

popop614

#11 h Jul 9, 2023, 8:00 PM • 1 Y

Y by centslordm

Define by $A'$ the reflection of $A$ over line $BC$ . Then observe that $\providecommand{\dang}{\measuredangle} \dang DBA = \dang DXA = \dang XAD = \dang XBD = \dang A'BD$ , so $BX$ is the reflection of line $AB$ over $BC$ . Hence $A'$ , $B$ , and $X$ are collinear. Similarly, $C$ , $Y$ , and $A'$ are collinear.

Now observe that $\frac{EF}{FD} = \frac{BF}{FC}$ so $EF \cdot FC = BF \cdot FD$ , implying that $AF$ is the radical axis of $(ABD)$ and $(ACE)$ . Moreover, $A'$ lies on $AF$ , so we obtain
$A'X \cdot A'B = A'A \cdot A'P = A'Y \cdot A'C,$ and so $XBCY$ is cyclic as desired.

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GrantStar

815 posts

GrantStar

#12 h Jul 9, 2023, 11:39 PM • 3 Y

Y by OronSH, centslordm, ihatemath123

Great problem!

Claim: The radical axis of $(ABD)$ and $(ACE)$ is line $AF$ .
Proof: From the parallel lines we have $\frac{FE}{FB}=\frac{FD}{FC}$ and rearranging we get $FE\cdot FC= FD\cdot FB$ as desired.

Now the arc midpoints give that $\measuredangle ABC=\measuredangle AXD=\measuredangle DAX=\measuredangle CBX$ so $BC$ is the exterior bisector of $\angle XBA$ and similarly it’s the exterior bisector of $\angle ACY$ .

Then, let $A’$ be the flection of $A$ across $BC$ . Then, $\angle A’BF=\angle CBA$ so $A’$ lies on $BX$ and similarly it lies on $CY$ . The result follows by the radical lemma as $A’$ lies on $AF$

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pi271828

3363 posts

pi271828

#13 h Jul 10, 2023, 12:50 AM

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Using the parallel lines it is clear that $AF$ is the radical axis of $(ABD)$ and $(ACE)$ . Let $XB \cap CY = Z$ . Notice that by the arc midpoints conditions, BC is the angle bisector of $\angle ABZ$ and $\angle ACZ$ . This implies $Z$ is the reflection of $A$ over $BC$ , and therefore $Z$ lies on $AF$ . This implies that $ZB \cdot ZX = ZY \cdot ZC$ , which implies that $BXCY$ is cyclic.

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JAnatolGT_00

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JAnatolGT_00

#14 h Jul 10, 2023, 12:37 PM

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Clearly $X,Y$ lie on external bisectors of angles $ABC,ACB$ respectively, so point $Z=BX\cap CY$ is a reflection of $A$ over $BC.$ By Thales $|EF|:|BF|=|FP|:|FA|=|FD|:|FC|\implies |EF|\cdot |FC|=|BF|\cdot |FD|,$ so $\overline{APFZ}$ is a radical axis of $\odot (ABD), \odot (ACE)$ implying the desired result by PoP.

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AlastorMoody

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AlastorMoody

#15 h Jul 10, 2023, 10:16 PM • 3 Y

Y by kamatadu, amar_04, Rounak_iitr

i think the main idea is similar, but nevertheless Solution

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Aiden-1089

277 posts

Aiden-1089

#45 h Mar 28, 2024, 3:47 AM

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Let $A'$ be the reflection of $A$ across $BC$ .
Note that $FB \cdot FD = FB \cdot (FC \cdot \frac{FP}{FA})=FC \cdot (FB \cdot \frac{FP}{FA}) = FC \cdot FE$ . So $F$ lies the radical axis of $(ABD)$ and $(ACE)$ . It follows that $AA'$ is the radical axis of the two circles.
Also, $\measuredangle XBD = \measuredangle XAD = \measuredangle DXA = \measuredangle DBA = \measuredangle A'BD$ , which implies that $X,B,A'$ are collinear.
Similarly, we have that $Y,C,A'$ are collinear. Since $BX$ and $CY$ intersect on the radical axis of $(ABD)$ and $(ACE)$ , we have that $B,C,X,Y$ are concyclic by the radical axis theorem.

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dstanz5

238 posts

dstanz5

#46 h Apr 10, 2024, 8:33 AM

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Is this possible with only angle chasing (no Radical Axis Theorem)?

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AngeloChu

470 posts

AngeloChu

#47 h Apr 11, 2024, 1:39 PM

Y by

let the other intersection of the two circles be $Q$
first, power of a point instantly yields that $Q$ lies on $AP$ , and therefore $AP$ is a radical axis of the two circles
let the extensions of $BX$ and $CY$ be $R$
since $DA=DX$ , $DBX=180-DBA$ and $DBA=DBR$
this yields $BA=BR$ , and similarly, we can get $CA=CR$
thus, $R$ is a reflection of $A$ over $BC$ , and lies on $AP$
therefore, $RX*RB=RC*RY$ by radical axis, and we are done

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Mogmog8

1080 posts

Mogmog8

#48 h Apr 28, 2024, 3:58 AM • 1 Y

Y by centslordm

Note $\triangle ABC\sim\triangle PED$ with $AB/PE=k$ . Then, $EF\cdot FC=\frac{BF}{k}\cdot kFD=BF\cdot FD$ so $F$ lies on the radical axis of $(ABD)$ and $(ACE)$ . Letting $Z=\overline{XB}\cap\overline{CY}$ , we have $\angle ZBC=180-\angle XBD=\angle ABD$ and similarly $\angle BCZ=\angle C$ so $Z$ is the reflection of $A$ over $F$ . Hence, $BXCY$ is cyclic by radical axis. $\square$

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alexgsi

139 posts

alexgsi

#49 h May 10, 2024, 9:32 PM

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dstanz5 wrote:

Is this possible with only angle chasing (no Radical Axis Theorem)?

Yes
sketch using only angle chasing

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sami1618

881 posts

sami1618

#50 h May 11, 2024, 3:45 PM

Y by

Sketch:
Claim: $F$ has equal power with both circles
Claim: $A'$ , the reflection of $A$ over $BC$ , lies on $BX$ and $CY$
Conclude by PoP.

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lnzhonglp

120 posts

lnzhonglp

#51 h Jun 18, 2024, 1:11 PM

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Let $Z = XB \cap YC$ . It suffices to prove that $Z$ lies on the radical axis of $(ABD)$ and $(ACE)$ . We have
$\measuredangle CBZ = \measuredangle DAX = \measuredangle AXD = \measuredangle ABC,$ so $BZ$ is the reflection of $BA$ over $BC$ . Similarly, $CZ$ is the reflection of $CA$ over $BC$ , so $Z$ is the reflection of $A$ over $BC$ and lies on line $AF$ . We have
$BF \cdot FD = BF \cdot FC \cdot \frac{PF}{AF} = EF \cdot FC,$ so line $AF$ is the radical axis of $(ABD)$ and $(ACE)$ , and $AF$ passes through $Z$ , as desired. $\square$

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Nuterrow

254 posts

Nuterrow

#52 h Sep 2, 2024, 9:37 AM

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First of all we claim that $(ABD) \cap (ACE) = T$ lies on $AF$ . This is true since from the parallel lines, we have that, $\frac{FD}{FC} = \frac{FP}{FA} = \frac{FE}{FB}$ which implies that $FE \times FC = FD \times FB$ which implies $F$ lies on the radical axis of $(ABD)$ and $(ACE)$ . Suppose $BX \cap CY = Z$ , now notice that if $Z$ lies on $AF$ , we will be done since then we would have $ZB \times ZX = ZT \times ZA = ZY \times ZC$ which implies the desired result. Now just notice that $\angle ABC = \angle ABD = \angle AXD = \angle XAD = \angle DBZ$ and similarly $\angle ACB = \angle ECZ$ , so $Z$ is just the reflection of $A$ across $BC$ and thus lies on $AF$ . $\blacksquare$

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pikapika007

297 posts

pikapika007

#53 h Nov 26, 2024, 9:56 PM

Y by

bored

Let $(ABD) \cap (ACE) = Z$ and let $A'$ be the reflection of $A$ over $BC$ .

Claim: $Z$ lies on $\overline{AF}$ (so that the radical axis of $(ABD)$ and $(ACE)$ is $\overline{AF}$ .)

Proof. Note that $\triangle ABC$ and $\triangle PED$ are homothetic with center $F$ . It follows that
$\frac{DF}{FE} = \frac{CF}{FB},$ or $CF \cdot FE = DF \cdot FB$ . Thus $F$ has equal power with respect to $(ABD)$ and $(ACE)$ , so $\overline{AF}$ is the radical axis of the two circles, as desired. $\square$

Finally, note that
$\measuredangle XBD = \measuredangle XAD = \measuredangle DXA = \measuredangle DBA = \measuredangle CBA = \measuredangle CBA'$ so $\overline{XBA'}$ collinear. Similarly $\overline{YCA'}$ collinear, hence $A'X \cdot A'B = A'Y \cdot A'C = A'Z \cdot A'A$ or $BCYZ$ cyclic.

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Thelink_20

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Thelink_20

#54 h Dec 24, 2024, 1:55 PM • 1 Y

Y by ohhh

No Bash? lol

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maths_enthusiast_0001

133 posts

maths_enthusiast_0001

#55 h Dec 30, 2024, 10:04 PM

Y by

Easy and a cute one :love:

Claim 1: $\color{blue}{AF,BX}$ and $\color{blue}{CY}$ are concurrent.
Proof: Note that $\square ABXD$ and $\square ACYE$ are cyclic quadrilaterals. As $DA=DX$ thus, both of these chords subtend equal angles at the center and hence at the circle. Thus, $\angle{ABD}=\angle{XBD}=\angle{XAD}=\alpha$ (say) and analogously, $\angle{ACE}=\angle{YCE}=\angle{EAY}=\beta$ (say). Also, $\angle{PED}=\alpha$ and $\angle{PDE}=\beta$ since, $PE \parallel AB$ and $PD \parallel AC$ . Let $Z$ be the reflection of $A$ over $BC$ . Then, $\angle{ZBC}=\alpha$ and $\angle{ZCB}=\beta$ implying, $\angle{ZBD}=\angle{XBD}$ and $\angle{ZCE}=\angle{YCE}$ . As $X,Y,Z$ all lie on the same side of $BC$ thus, $AF,BX$ and $CY$ are concurrent at $Z$ as desired. $\blacksquare$
Claim 2: $\color{blue}{AZ}$ is the radical axis of $\color{blue}{(ABXD)}$ and $\color{blue}{(ACYE)}$ .
Proof: Let $(ABXD) \cap AZ=O$ . Then $\angle{AOD}=\angle{ABD}=\alpha$ but we also have $\angle{PED}=\alpha$ thus, $\angle{PED}=\angle{POD}$ implying $P,E,O,D$ lie on a circle. Thus, $\angle{POE}=\angle{PDE}=\beta$ but $\angle{ACE}=\alpha$ thus, $\angle{ACE}=\angle{AOE}$ implying $A,C,Y,O,E$ lie on a circle. Thus we have $(ABXOD)$ and $(ACYOE)$ . Clearly $AO$ (or $AZ$ ) is the radical axis of these circles as desired. $\blacksquare$
Claim 3: $\color{blue}{B,C,X,Y}$ are concyclic.
Proof: Call $(ABXOD)$ and $(ACYOE)$ as $\Gamma_{1}$ and $\Gamma_{2}$ respectively. Computing powers of $Z$ we get,
$Pow_{(\Gamma_{1})}Z=Pow_{(\Gamma_{2})}{Z} \implies \boxed{ZX.ZB=ZY.ZC}$ implying $B,C,X,Y$ are concylic as desired. $\blacksquare$ ( $\mathcal{QED}$ )
Remark: With trivial angle chasing one can also show that $O$ is the circumcenter of triangle $XYZ$ .

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Bluesoul

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Bluesoul

#56 h Feb 18, 2025, 3:13 AM

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By the parallel condition we have $\frac{FE}{FB}=\frac{FD}{FC}=\frac{FP}{FA}$ so $BF\cdot FD=CF\cdot FE$ so $AF$ lies on the radical axis of $(ABD), (ACE)$

Then by the equal length condition we have $\angle{AXD}=\angle{ABD}=\angle{XAD}=\angle{DBN}; \angle{EAY}=\angle{EYA}=\angle{ECY}=\angle{ECA}$ which implies $XB\cap CY$ is the reflection of $A$ about $BC$ which means $XB\cap CY$ lies on the radical axis of $(ABD), (ACE)$ and we are done.

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Retemoeg

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Retemoeg

#57 h Feb 18, 2025, 4:41 PM

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Let $K$ be the intersection of $BX$ and $CY$ . Note that, since $DA = DX$ and $EA = EY$ , we should have that $BD$ and $CF$ bisects $\angle ABX$ and $\angle ACY$ respectively, thus $A$ and $K$ are symmetrical wrt $BC$ , implying that $A, F, K$ are collinear. Note that:
$\cfrac{FD}{FC} = \cfrac{FP}{FA} = \cfrac{FE}{FB} \implies FD\cdot FB = FE\cdot FC$ So if $L$ is the second intersection of $AF$ with $(ABD)$ then $FL\cdot FA = FD\cdot FB = FE\cdot FC$ . So $L$ also lies on $(ACE)$ .
Now, $A, F, L, K$ are collinear so $KX\cdot KB = KL\cdot KA = KY\cdot KC$ by power of a point, implying that $BXYC$ is cyclic and we are done.

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Tony_stark0094

45 posts

Tony_stark0094

#58 h Mar 5, 2025, 2:33 PM

Y by

easy geo:
since EP is parallel to AB we can deduce AF to be the radax of (ABD) and (AEC)
now let BX and CY meet at T
it is enough to prove AT is perpendicular to BC
$\angle ECY =\angle EAY =\angle EYA =\angle ECA$ and similarly i have shown $\angle ABD=\angle TBD$
so T is the reflection of A about BC hence we are done

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Reason: rposwst

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Ilikeminecraft

330 posts

Ilikeminecraft

#59 h Mar 11, 2025, 10:13 PM

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Construct point $A’$ so that it is the reflection of $A$ across $BC.$ Observe that $\angle XBC = 180 - \angle XAD = 180 - \angle AXD = 180 - \angle ABD = 180 - \angle DBA’,$ so $X, B, A’$ are collinear. Similarly, $A’, C, Y$ are collinear.
Define $T = (PED)\cap AF.$ We get that $\angle ETP = \angle PDE = \angle ACB,$ and so $(AHTC)$ is cyclic. Similarly, $(AXBT)$ is cyclic.
Hence, the radax of $(AXB), (ACH)$ is $AG = AF,$ which also passes through $E.$ Thus, $A’B \cdot A’X = A’G\cdot A’A = A’C \cdot A’Y,$ which finishes.

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