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 $EP=EQ.$

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.

An "extra" question: Prove that $EN$ bisects $\angle{CND}$.

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.

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.

$K$ 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.

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

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 $

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. $\blacksquare$

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.

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

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

Yet additional requirement: the quadrilateral $ANBE$ is harmonic! Best regards, sunken rock

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 $EP=EQ.$ When I tried to do this problem I did exactly this except I didn't see that PM=QM...

Can one prove this with inversion?

Sorry to revive but I thought it was cool that once you look at the homothety and get that M is the midpoint of PQ if you angle chase a bit you get that AEBM is a kite and therefore EM is perpendicular to AB and therefore to CD at M and your done

Dear Mathlinkers, just to mention that a enterely synthetic proof without calculation, angles, transformation is possible... Sincerely Jean-Louis

Let $NM \cap AB=X$ since $X$ lies on the radical axis of $G_1$ and $G_2$, $XA^2=XB^2$ which means $XA=XB$. By similarity $QM=PM$. Since $\angle BAM = \angle MCE= \angle BAE$ and $\angle MBA = \angle EDC= \angle EBA$ quadrilateral $AEBM$ is a kite. So $\angle EMC =90^{\circ}$. So $QEP$ is isosceles finishing the problem.

Note $MN$ bisects $AB$ by radical axis. Because $AB \parallel PQ$, there exists a homothety mapping $AB$ to $PQ$ centered at $N$ which also maps the midpoint of $AB$ to $M$. Hence $M$ is the midpoint of $PQ$. The problem reduces to proving $EM \perp CD$, or $EM \perp AB$. At this point, points $M$, $N$, $P$, and $Q$ are irrelevant. We can angle chase to get \[\angle BAM = \angle AMC = \angle ACM = \angle EAB,\]\[\angle ABM = \angle BMD = \angle BDM = \angle EBA.\] Consequently, $EAMB$ is a kite, giving our desired conclusion.

First, notice that we have $$\angle CMA = \angle BAM = \angle ACM = \angle EAB$$and similarly $$\angle BMD = \angle ABM = \angle BDM = \angle EBA$$Let $X$ be the intersection of $EM$ and $AB$. Now, notice that $$CA = AM = AE \text{ and } BM = BD = EB$$And by congruency we have that $$\angle AXE = \angle AXM = 90^\circ$$Thus, $\angle CME = 90^\circ$ as well using Parallel Lines. Now, notice that $MN$ bisects $AB$ and thus using the parallel lines, $MN$ bisects $PQ$ as well. Thus, $PM=MQ$. Therefore, $\triangle EPM \cong \triangle EQM$ which gives us that $EP=EQ$ as required.

Wow I never posted soln for this problem. $\angle EAB=\angle ECD=\angle ACM=\angle BAM$ and similarly $\angle EBA = \angle ABM \implies AEBM$ is kite $\implies EM\perp AB\implies EM \perp CD$. Now $M$ is midpoint of $PQ$ due to homothety at $N$ mapping $AB\mapsto PQ$. Thus $\triangle EPQ$ is isosceles $\implies EP=EQ$.

Let $\overline{EN} \cap \overline{AB} = K$. $\newline$ Notice that \[\angle{ACM} = \angle{AMC} = \angle{MAB}\]and similarly \[\angle{MBA} = \angle{EBA}\]so $\triangle EBA \cong \triangle MBA$ are directly similar $\implies EBMA$ is a kite. This implies that $EN \perp AB$. Since $K$ lies on the radical axis of $G_1$ and $G_2$, $KA = KB$ so the homothety sending $AB$ to $PQ$ also sends $K$ to $M$. $\newline$ Since the $E$-altitude and the $E$-median of $\triangle EPQ$ coincide(they intersect $PQ$ at $M$), $\triangle EPQ$ is isosceles, specifically $EP = EQ$, which finishes. $\blacksquare$

My first problem after 1 month of INMO It is well known that $M$ is altitude. Therefore $ EM\perp CD$. Also well known that $NM$ bisects $AB$. Therfore we get $MP=MQ.$ Now we get our desired result.

Let $O$ be the center of $G_2$. Claim 1: $EM \parallel BO$ Proof. Let $EM$ intersect $G_2$ at a second point $X \neq M$. First, we have $AB \parallel CD$, and angle chasing leads to $$\angle BMD = \angle ABM = \angle BDM$$It follows that $$\angle MXD = m\overset{\Large\frown}{BD} = \angle BQD$$and the claim follows. $\blacksquare$ Since we have $OB \perp AB$ and $AB \parallel CD$, we obtain $EM \perp CD$. Lemma (Trapezoid Lemma): In trapezoid $ABCD$ ($AB \parallel CD$), let $N$ be the midpoint of $CD$, and $X$ be the intersection of $AD$ and $BC$. The intersection $M$ of $XN$ and $AB$ is the midpoint of $AB$. Proof. By similar triangles, we obtain $$\dfrac{AM}{DN} = \dfrac{XM}{XN} = \dfrac{MB}{NC}$$However, $N$ is the midpoint of $CD$, so it follows that $AM=MB$, or $M$ is the midpoint of $AB$. $\blacksquare$ Returning to the original problem, let $K = MN \cap AB$. Since $K$ lies on the radical axis of $G_1$ and $G_2$, we have $\text{Pow}_{G_1} (K) = \text{Pow}_{G_2} (K)$, which implies that $KA=KB$, or $K$ is the midpoint of $AB$. Applying the lemma above on trapezoid $ABQP$, we have $M$ is the midpoint $PQ$. Thus, $\Delta EPQ$ is isoceles with $EP=EQ$, as desired. $\blacksquare$

Fun problem, and my first ever post on AoPS about an actual oly problem! Diagram attached. Claim 1: $MP = MQ$ Proof: Call $K$ the point where radical axis $MN$ intersects line $AB$. We understand that as $K$ lies upon the radical axis of $G_1$ and $G_2$. This means that the power of $K$ to both circles is equal. $(KA)^2 = (KB)^2$ $\rightarrow KA = KB$ Therefore, $K$ is the midpoint of $AB$. Now, as $\triangle NAB \sim \triangle NPQ$, we also know that $M$ must be the midpoint of PQ as well, giving us this step of the proof. Claim 2: $EM \perp CD$ Proof: First, we prove that $\triangle EAB \sim \triangle ECD$ by a scale factor of 2. We can easily see that the two triangles are similar. As to their scale factor, we can see that from the Perpendicular Bisector of a Chord Theorem, segment $CD$ is twice as long as segment $AB$. Now, we use this fact to see that, after constructing the perpendicular from $E$ to $CD$, the foot of the altitude must be $M$ by similar triangles. Finish: Now, we have the following facts: $MP = MQ$ and $EM \perp CD$. This easily gives us that $EP = EQ$ by congruence.

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 ?

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. $\blacksquare$

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 $EP = EQ$.

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 $EP=EQ.$

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.

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

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