Two circles, a tangent line and a parallel

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#1 h Oct 24, 2005, 10:15 AM • 19 Y

Y by Davi-8191, OlympusHero, FaThEr-SqUiRrEl, lc426, centslordm, Adventure10, Designerd, jhu08, megarnie, Numbertheorydog, HWenslawski, Kanimet0, ImSh95, Zhaom, Mango247, Rounak_iitr, Math_.only., ItsBesi, cubres

Two circles $G_1$ and $G_2$ intersect at two points $M$ and $N$ . Let $AB$ be the line tangent to these circles at $A$ and $B$ , respectively, so that $M$ lies closer to $AB$ than $N$ . Let $CD$ be the line parallel to $AB$ and passing through the point $M$ , with $C$ on $G_1$ and $D$ on $G_2$ . Lines $AC$ and $BD$ meet at $E$ ; lines $AN$ and $CD$ meet at $P$ ; lines $BN$ and $CD$ meet at $Q$ . Show that $EP = EQ$ .

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arqady

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arqady

#2 h Oct 24, 2005, 10:58 AM • 29 Y

Y by jam10307, shiningsunnyday, claserken, shawnee03, futurestar, Richangles, myh2910, OlympusHero, User582032, starchan, sotpidot, FaThEr-SqUiRrEl, centslordm, Adventure10, megarnie, asdf334, Numbertheorydog, HWenslawski, rayfish, Kanimet0, ImSh95, Mango247, CoC_Ali, Rounak_iitr, Math_.only., ehuseyinyigit, cubres, and 2 other users

let MN and AB intesect at K. Then $KA^2=KM*KN=KB^2.$ Hence KA=KB.
CD||AB. Hence PM=QM.
$\angle AEB=\angle ECD=\angle MAB,$ $\angle EBA=\angle EDC=\angle MBA.$
hence $\triangle AMB$ and $\triangle AEB$ congruity.
Hence $EM\perp AB.$ Hence $EM\perp PQ.$
Hence $EP=EQ.$

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yetti

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yetti

#3 h Oct 30, 2005, 11:05 PM • 10 Y

Y by JasperL, FaThEr-SqUiRrEl, centslordm, Adventure10, Designerd, jhu08, ImSh95, Rounak_iitr, cubres, and 1 other user

The radical axis MN of the circles $(G_1), (G_2)$ cuts the tangent length segment AB at its midpoint C. Since $PQ \parallel AB$ , the triangles $\triangle ABN \sim \triangle PQN$ are centrally similar with similarity center N. Hence, M is the midpoint of the segment PQ. The radii $G_1A, G_2B$ of the circles $(G_1), (G_2)$ are perpendicular to the tangent AB, i.e, also to the chords AM, DM. Hence they cut these chords at their midpoints, which means that the triangles $\triangle ACM, \triangle BDM$ are isosceles and the angles $\angle ACM = \angle AMC, \angle BDM = \angle BMD$ are equal. Since the lines $AB \parallel CD$ are parallel and $M \in CD$ , the angles $\angle MAB = \angle AMC = \angle ACM = \angle EAB$ are equal and similarly, the angles $\angle MBA = \angle BMD = \angle BDM = \angle EBA$ are also equal. Thus the diagonal AB of the quadrilateral AMBE bisects its opposite angles at the vertices A, B, which implies that this quadrilateral is a kite with AE = AM, BE = BM and and its diagonals $AB \perp EM$ are perpendicular to each other. Consequently, the lines $EM \perp CD \equiv PQ$ are also perpendicular to each other. Since M is the midpoint of the segment PQ, the line EM is the perpendicular bisector of this segment. It follows that the triangle $\triangle EPQ$ is isosceles with EP = EQ.

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Arne

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Arne

#4 h Oct 30, 2005, 11:10 PM • 8 Y

Y by mihajlon, OlympusHero, FaThEr-SqUiRrEl, Adventure10, ImSh95, cubres, and 2 other users

An "extra" question:

Prove that $EN$ bisects $\angle{CND}$ .

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Zhero

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Zhero

#5 h Apr 8, 2010, 1:28 AM • 5 Y

Y by FaThEr-SqUiRrEl, ImSh95, Adventure10, Mango247, cubres

Since $AB$ is the common tangent of $G_1$ and $G_2$ and $MN$ is the radical axis of $G_1$ and $G_2$ , $MN$ bisects $AB$ . A homothety centered at $N$ that maps $A$ to $P$ and $B$ to $Q$ therefore maps the midpoint of $AB$ to $PQ$ , so we have that $M$ is the midpoint of $PQ$ .

It is sufficient to show that $EM \perp CD$ . Let $R$ and $S$ be the projections of $A$ and $B$ onto $CD$ , respectively; since $AB || CD$ , $AR = BS$ . We first note that $AC = AM$ , since $m \angle AMC = m \angle EAB = m \angle ACM$ (as $AB$ is tangent to $G_1$ ). We can similarly deduce that $BM = BD$ .

It also follows from this that $CM = 2CR$ and $DM = 2DS$ . Let $E'$ be the point such that $E'M \perp CD$ and $E'M = 2AR = 2BS$ . A homothety centered at $C$ that maps $R$ to $M$ must therefore map $A$ to $E'$ , and a homothety centered at $D$ must therefore map $B$ to $E'$ . But this means that $E'$ lies on both $AC$ and $BD$ , implying that $E' = E$ . It follows that $EM \perp CD$ , so we are done.

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Pikachu!!!

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Pikachu!!!

#6 h Apr 11, 2010, 4:30 AM • 6 Y

Y by S117, FaThEr-SqUiRrEl, ImSh95, Adventure10, Mango247, cubres

Arne wrote:

An "extra" question:

Prove that $EN$ bisects $\angle{CND}$ .

This extra question is also beautiful one.

First, from what we know Let $\angle{ANM}=\angle{BAM}=\angle{BAE}=x$ and $\angle{BNM}=\angle{ABM}=\angle{ABE}=y$

Therefore, $\angle{AEB}+\angle{ANB}= (\pi-x-y)+(x+y)=\pi$ $\longrightarrow$ $A,N,B,E$ is concyclic.

It follows that $\angle{BNE}=\angle{BAE}=x$ and it is known that $\angle{BND}=\angle{BNM}=y$
We get $\angle{DNE}=x+y$

Similarly, we also have $\angle{CNE}=x+y$

This is what we want to prove.

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arshakus

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arshakus

#7 h Jun 13, 2011, 6:50 AM • 5 Y

Y by FaThEr-SqUiRrEl, ImSh95, Adventure10, Mango247, cubres

$K$ is the midpint of $AB$ . $NM$ is radical axis for $G_1,G_2$ circles. also we know that radical axis passes through the midpoints of comment tangents. Thus $N,M,K$ lies on the same line. $PQ||AB=>M$ is the midpoint of $PQ$ .
Also it is easy to note this number of equations.
$\angle{PCA}=\angle{ANM}=\angle{BAE}=\angle{MAB}=\angle{PMA}=\angle{CNA}=\angle{ENB}$ [*] the last equation follows from the fact that $AENB$ is cyclic.
Similarly we can get in the opposite side. Thus $\triangle {AMB}=\triangle {AEB}=>EM\perp PQ=>EP=EQ$
@Arne
from [*] it is easy to note it.

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arshakus

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arshakus

#8 h Jun 13, 2011, 6:56 AM • 5 Y

Y by FaThEr-SqUiRrEl, ImSh95, Adventure10, Mango247, cubres

One more "extra" question!)
Prove that $2AB=CD$

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siavosh

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siavosh

#9 h Mar 3, 2012, 7:14 PM • 5 Y

Y by FaThEr-SqUiRrEl, ImSh95, Adventure10, Mango247, cubres

arshakus wrote:

One more "extra" question!)
Prove that $2AB=CD$

In your solution we prove that $\triangle{AMB}=\triangle{AEB}=> EB=BM=BD$
and $CD\parallel AB => \frac{EB}{ED}=\frac{AB}{CD}=2$

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JSGandora

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JSGandora

#10 h Apr 17, 2013, 1:59 AM • 6 Y

Y by FaThEr-SqUiRrEl, ImSh95, Adventure10, Mango247, Rounak_iitr, cubres

Let $\angle ACM=\alpha$ , then since $AB||CD$ , we have $\angle EAB=\angle ACM=\alpha$ and since $AB$ is tangent to circle $G_1$ , we have $\angle BAM=\angle ACM=\alpha$ . Thus $\angle EAB=\angle BAM=\alpha$ . Similarly, if we let $\angle BDM=\beta$ , we find $\angle EBA=\angle ABM=\beta$ .

Now let the intersection of $EM$ and $AB$ be $F$ , then $\angle AFE=90^\circ$ and $\angle CME=\angle AFE=90^\circ$ and therefore proving $EP=EQ$ is equivalent to proving $PM=QM$ since $EM\perp PQ$ .

Let the intersection of ray $NM$ and $AB$ be point $G$ , then $GA=GB$ because $G$ is on the radical axis of the two circles and thus $GA^2=GB^2\implies GA=GB$ . Additionally, since $AB||CD$ , we have $\triangle AGN~\triangle PMN$ and $\triangle BGN~\triangle QMN$ and therefore $PM=PQ$ as desired. $\blacksquare$

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BBAI

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BBAI

#11 h May 3, 2013, 9:14 AM • 4 Y

Y by FaThEr-SqUiRrEl, ImSh95, Adventure10, cubres

As the tangent is parallel to the chords So it follows that $\triangle ACM$ and $\triangle BDM$ are isosceles and $EA=AM=AC$ so $EM$ is perpendicular to $PQ$ . Define $Z=MN \cap AB$ . $Z$ lies on radical axis ,so $Z$ is the midpoint of $AB$ . And hence $M$ is the midpoint of $PQ$ . So done.

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subham1729

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#12 h May 3, 2013, 11:38 AM • 4 Y

Y by FaThEr-SqUiRrEl, ImSh95, Adventure10, cubres

OMG!! got a proof without bashing :O - if $MN$ meets $AB$ at $K$ then $AK^2=BK^2\implies AK=KB\implies PM=MQ\implies \angle{CMN}=\frac {\pi}{2}\implies EP=EQ$

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BBAI

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BBAI

#13 h May 3, 2013, 12:01 PM • 4 Y

Y by FaThEr-SqUiRrEl, ImSh95, Adventure10, cubres

I think $E,M,N$ are not at all collinear.
I have drawn a ggb diagram. and checked it.

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sunken rock

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sunken rock

#14 h May 6, 2013, 9:01 PM • 6 Y

Y by FaThEr-SqUiRrEl, ImSh95, Adventure10, Mango247, cubres, and 1 other user

Yet additional requirement:

the quadrilateral $ANBE$ is harmonic!

Best regards,
sunken rock

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exmath89

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exmath89

#15 h Jun 15, 2013, 10:02 PM • 7 Y

Y by Math-Star., FaThEr-SqUiRrEl, ImSh95, Adventure10, Rounak_iitr, cubres, and 1 other user

Solution

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Math_.only.

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Math_.only.

#96 h Aug 1, 2024, 8:22 AM • 1 Y

Y by cubres

arqady wrote:

let MN and AB intesect at K. Then $KA^2=KM*KN=KB^2.$ Hence KA=KB.
CD||AB. Hence PM=QM.
$\angle AEB=\angle ECD=\angle MAB,$ $\angle EBA=\angle EDC=\angle MBA.$
hence $\triangle AMB$ and $\triangle AEB$ congruity.
Hence $EM\perp AB.$ Hence $EM\perp PQ.$
Hence $EP=EQ.$

Why angles AEB and ECD are equal ?

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MagicalToaster53

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MagicalToaster53

#97 h Sep 15, 2024, 12:31 AM • 1 Y

Y by cubres

Notice that $MN$ bisects $\overline{AB},$ as the point where they intersect has equivalent powers to the respective circles. From this we discover that $M$ is the midpoint of $\overline{PQ}$ due to the homothety at $N$ mapping $\overline{PQ} \mapsto \overline{AB}.$ It then suffices to show $EM \perp PQ$ .

Claim: $\triangle EAB \cong \triangle MAB.$
Proof: Observe that $AB$ bisects $\angle EAM:$ $\angle EAB = \angle ECM = \angle ACM = \angle BAM.$ Similarly by symmetry, we find that $AB$ bisects $\angle EBM.$ Then by SAS, we conclude that $\triangle EAB \cong \triangle MAB,$ as was to be shown. $\square$

Now $EAMB$ is a kite, so that $EM \perp AB \implies EM \perp PQ,$ as $AB \parallel PQ$ . Hence $\triangle EPQ$ is isosceles, as desired. $\blacksquare$

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lnzhonglp

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lnzhonglp

#98 h Nov 27, 2024, 4:43 AM • 1 Y

Y by cubres

Observe that $\measuredangle EBA = \measuredangle EDC = \measuredangle ABM$ and $\measuredangle EAB = \measuredangle ECM = \measuredangle MAB,$ so $\triangle EAB \cong \triangle MAB$ and $EM \perp AB$ . From radical axis we see that $MN$ bisects $AB$ , so by homothety at $N$ we find that $M$ is the midpoint of $PQ$ . Then since $EM \perp PQ$ it follows that $EP = EQ$ .

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Likeminded2017

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Likeminded2017

#99 h Dec 24, 2024, 2:01 PM • 1 Y

Y by cubres

Let $MN$ and $AB$ intersect at $K.$ As $K$ lies on the radical axis we have $KA=KB.$ Taking the homothety centered at $N$ that maps $M$ to $K$ we find that $MP=MQ.$ Then $\angle BAM=\angle MCA=\angle BAE$ and similarly $\angle EBA=\angle ABM$ so by ASA congruence $\triangle EAB \cong \triangle MAB.$ Thus $EM$ is perpendicular to $AB$ and thus $PQ$ so $EP=EQ.$

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ezpotd

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ezpotd

#100 h Jan 4, 2025, 12:27 PM • 1 Y

Y by cubres

Letting $AB$ meet $MN$ at $K$ , we see $KA^2 = KM \cdot KN = KB^2$ , so $K$ is the midpoint of $AB$ , taking the homothety from $AB$ to $PQ$ this gives $MP = MQ$ , so it suffices to prove $EM \perp PQ$ , or $EM \perp AB$ . However, angle chasing gives $\angle EAB = \angle ECM = \angle AMC = \angle BAM, \angle EBA = \angle EDM = \angle BMD = \angle ABM$ , so $\triangle EAB \cong \angle MAB$ , so $E,M$ are reflections about $AB$ , done.

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optimusprime154

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optimusprime154

#101 h Jan 4, 2025, 5:36 PM • 2 Y

Y by SorPEEK, cubres

let $Z$ = $MN \cap AB$ from PoP we know $ZB^2 = ZM * ZN$ and $ZA^2 = ZM * ZN$ which means $ZA = ZB$ and since $PQBA$ is a trapezoid, $PM = MQ$
now we prove $EM \perp CD$ and we're done. label $\angle ACM = x, \angle BDM = y$ we know from $AB \parallel CD$ that $\angle EAB = x$ and $\angle EDC = y$
we also know from $AB$ being tangent that $\angle MBA\ = y$ and $\angle MAB = x$ the above facts indicate the similarity of $\triangle MAB , \triangle EAB$ so $EM \perp AB$ => $EM \perp CD$ and we're done

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Maximilian113

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Maximilian113

#102 h Jan 10, 2025, 12:54 AM • 1 Y

Y by cubres

Extend $MN$ to intersect $AB$ at $K,$ then clearly $AK=BK.$ Therefore $PM=MQ$ by the parallel lines.

Now, notice that $\angle EAB = \angle ACM = \angle BAM = \angle AMC.$ Therefore $AB$ bisects $\angle EAM$ and $AC=AM.$ Similarly $AB$ bisects $EBM$ and $BD=BM.$ Therefore, $\triangle EAB \cong \triangle MAB$ so $A, B$ are the circumcenters of $\triangle EMC, \triangle EMD$ respectively so by Thales' Theorem it follows that $\angle EMP = \angle EMQ = 90^\circ.$ Hence $EM$ is the perpendicular bisector of $PQ,$ and we are done. QED

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ehuseyinyigit

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ehuseyinyigit

#104 h Feb 10, 2025, 8:37 PM • 1 Y

Y by cubres

WELL-KNOWN. Since $AB\parallel CD$ , we have $\angle ACM=\angle AMC=\angle MAB=\angle BAE$ Thus, $AC=AM$ and $AB$ bisects $\angle EAM$ . Similarly $BD=BM$ and $AB$ bisects $\angle EBM$ hence $ABME$ is a kite $\Rightarrow AE=AM=AC$ implying $EM\perp CD$ . On the other hand, for point $F$ being intersection of lines $AB=MN$ , $AF=BF$ implies $PM=MQ$ . We obtain $EP=EQ$ .

WELL-KNOWN. We will show $EN$ bisects $\angle CND$ . This is true because $\angle ENB=\angle EAB=\angle ENC$ and $\angle ANE=\angle ABE=\angle BND$ . Thus, $EN$ bisects $\angle CND$ .

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mathwiz_1207

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mathwiz_1207

#105 h Feb 23, 2025, 11:52 PM • 1 Y

Y by cubres

We have
$\measuredangle EAB = \measuredangle ACM = \measuredangle ANM = \measuredangle BAM = \measuredangle CMA$ $\measuredangle EBA = \measuredangle BDM = \measuredangle BNM = \measuredangle ABM = \measuredangle DMB$ , implying that $\triangle AMB \cong \triangle AEB$ , and that $\triangle AMC, \triangle BDM$ are isosceles. Therefore,
$AE = AM = AC$ so $A$ is the center of $(EMC)$ , and similarly $B$ is the cetner of $(EMD)$ . Thus, $EM \perp CD$ . Now, let $L = MN \cap AB$ , by Radical Axis,
$LA = LB$ and since $\triangle NPQ \sim \triangle NAB$ , we have $MP = MQ \implies EQ = EP$ , so we are done.

This post has been edited 1 time. Last edited by mathwiz_1207, Feb 23, 2025, 11:53 PM
Reason: forgot to add last line

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study1126

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study1126

#106 h Mar 8, 2025, 6:51 AM • 1 Y

Y by cubres

All angles will be directed in this solution.

First, $\angle EAB=\angle ECB=\angle BAM$ , since $AB||CD$ and $AB$ is tangent to the two circles. Similarly, $\angle EBA=\angle ABM$ . Combined with $AB=AB$ gives $\triangle EAB\cong \triangle MAB$ .

This implies $EAMB$ is a kite, so $EM\perp AB$ , which means $EM\perp CD$ . Since $M$ is on the radical axis, extending $NM$ to $X$ on $AB$ gives $X$ is the midpoint of $AB$ . Homothety shows $M$ is the midpoint of $PQ$ . Thus, $MP=MQ$ , so by congruent triangle $\triangle EPM\cong \triangle EQM$ , $EP=EQ$ , as desired.

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joshualiu315

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joshualiu315

#107 h Mar 15, 2025, 12:40 AM • 1 Y

Y by cubres

It is well-known that $\overline{MN}$ bisects $\overline{AB}$ , and since $\overline{AB} \parallel \overline{CD}$ , we also know that $\overline{MN}$ bisects $\overline{PQ}$ . This implies $M$ is the midpoint of $\overline{PQ}$ .

We have $\angle EAB = \angle ACM = \angle MAB$ and similarly, $\angle EBA = \angle BDM = \angle MBA$ . This means $\triangle EAB \cong \triangle MAB$ , which implies that $\overline{EM} \perp \overline{AB}$ . Because $\overline{AB} \parallel \overline{CD}$ , we have $\overline{EM} \perp \overline{PQ}$ , so $\overline{EM}$ is the perpendicular bisector of $\overline{PQ}$ . Thus, $EP=EQ$ , as desired. $\blacksquare$

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LeYohan

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LeYohan

#108 h Apr 4, 2025, 12:10 AM • 1 Y

Y by cubres

First IMO #1 solve and it's geo!!

Claim: $PM = MQ$ .
Let $NM$ intersect $AB$ at $X$ . It's well known that the radical axis intersects the common tangent, hence $AX = XB$ . By the Thales Theorem, we inmediately get the result. $\square$

Now doing some angle chase using the fact that $CD \parallel AB$ and $AB$ is tangent to both circles, we get:
$\angle EBA = \angle BDM = \angle BNM = \angle ABM = \angle BMD$ , and $\angle EAB = \angle ACM = \angle ANM = \angle BAM = \angle AMC$ . As a result, $\triangle ACM$ and $\triangle BMC$ are isosceles. Furthermore, $\triangle MAB \cong \triangle EAB \implies EB = BD, EA = AC$ . This means that $\angle EMC = \angle EMD = 90$ , and notice that we originally proved that $PM = MQ$ , so $\triangle EPQ$ is isosceles and we're done. $\square$

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Avron

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Avron

#109 h Apr 7, 2025, 6:45 PM • 1 Y

Y by cubres

$MN$ is the radical axis of $G_1,G_2$ so it bisects $AB$ so we also know that $M$ is the midpoint of $PQ$ , thus it is enough to prove $ME\perp CD$ .
Now notice that $\angle MAB=\angle ACM = \angle EAB$ and similarly $\angle MBA = \angle ABE$ so $EAB \cong MAB$ and we're done

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zuat.e

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zuat.e

#110 h Apr 23, 2025, 4:18 PM • 1 Y

Y by cubres

Easy angle chasing yields the cyclicity of $(BEAN)$ . We claim the following:

Claim: $EA=AM=AC$ and similarly $EB=BM=BD$
Proof: Let $O_1$ be the center of $G_1$ . It is clear that $O_1A$ is the side bisector of $MC$ , hence $AM=AC$ .
Furthermore, note that $\measuredangle BAE=\measuredangle DCE= \measuredangle BMC=\measuredangle MAB$ and similarly $\measuredangle EBA =\measuredangle ABM$ , hence $\triangle EAB\cong MAB$ (because they also share $AB$ ).

It now follows that $AC=AM=AE$ and analogously $BD=BM=BE$
Finally, as $\measuredangle MCA= \measuredangle MNA=\measuredangle ANC$ , we have $AP\cdot AN=AC^2=AE^2$ , hence $\measuredangle CEP =\measuredangle ENA=\measuredangle EBA=\measuredangle EDC$ and similarly $\measuredangle QED=\measuredangle PCE$ , yielding $\measuredangle CPE =\measuredangle EQD$ , as desired.