thegreatp.d wrote:

How could you assume that $OP$ is the common radius. It can be like that $\alpha$ doesn't pass through center of $\omega$

I assumed that $O$ & $P$ are the center-and $\alpha$ also goes through the center of $\omega$. I mean the two circles are equal

thegreatp.d wrote: I assumed that $O$ & $P$ are the center-and $\alpha$ also goes through the center of $\omega$. I mean the two circles are equal A speacial case only.Not a full proof.

Any idea please?

Olympus_mountaineer wrote: Let $\alpha$ and $\omega$ be two circles such that $\omega$ goes through the center of $\alpha$.$\omega$ intersects $\alpha$ at $A$ and $B$.Let $P$ any point on the circumference $\omega$.The lines $PA$ and $PB$ intersects $\alpha$ again at $E$ and $F$ respectively.Prove that $AB=EF$. Let $O$ be the circumcenter of $\alpha$. $OP\cap AB=Q$. Observe that $O$ is the midarc of $\overarc{AB}$; so $OP\cdot OQ=OA^2=OE^2=OB^2=OF^2$. Hence, $\angle OAQ=\angle OPA=\angle OPE=\angle OEQ$. In other words, $\square OQAE$ is concyclic. Analogously, $\square OQFB$ is concyclic too! Now, by basic angle-chasing, we obtain that $E,F,Q$ are collinear and $\angle EOF=\angle AOB$. Done

Let $O$ be the center of $\alpha$, hence $PBOA$ is cyclic $\angle AOB = 180º-\angle BPA$. For other side: $\angle AOB = 2\angle AEB$ (central angle), look to $\triangle PBE$, $\angle AEB = \frac{180º-\angle BPA}{2}$, then $PB = PE$, by power of point: $PF.PB=PA.PE \implies FB=AE$, thus $FABE$ is isosceles trapezoid, and $FE=AB$. \box$

Let $O$ be center of $\alpha$ then $PO$ is internal bisector of $\widehat{APB}$ But: $OE$ = $OF$ so: $P$, $E$, $O$, $F$ lie on a circle Hence: $\widehat{EOF}$ = $180^o$ $-$ $\widehat{EPF}$ = $180^o$ $-$ $\widehat{APB}$ = $\widehat{AOB}$ Therefore: $\stackrel\frown{EF}$ = $\stackrel\frown{AB}$ or $EF$ = $AB$

This is my solution. Please correct me if I made some mistakes. [asy][asy] size(10cm,0); defaultpen(fontsize(10pt)); pair A = (0.8,1.833); pair B = (0.8, -1.833); pair P = (4,2); pair E = (-0.986, 1.74); pair F = (1.946, -0.46); pair O = (0,0); draw(circle((0,0), 2)); draw(circle((2.5,0), 2.5)); draw(B--P--E); draw(E--F--A--cycle); draw(A--B); draw(O--P); draw(O--A,red); draw(O--B,red); dot("$A$",A,2*N); dot("$B$",B,2*S); dot("$P$",P,2*right); dot("$E$",E,2*W); dot("$F$",F,SE); dot("$O$",O,left); [/asy][/asy] Since we have $OA=OB$ then $\angle OAB=\angle OBA$, which implies that $\angle OPA=\angle OPB$. And hence, $\triangle APF$ is isosceles. Then \[PA=PF \quad \text{and}\quad \angle EAF=\angle AFB \tag{1}\]From Power of Point Theorem which respect of the point P, we obtain that $PA\cdot PE=PF\cdot PB$. Using $(1)$ , it follows that \[AE=FB\tag{2}\]Combining the relation $(1)$ and $(2)$ , yields \[\triangle AEF\cong\triangle FBA.\]Therefore we get $AB=EF$ as desired.

I haven't seen this solution yet, so there's either something wrong with it or it's too simple to be noticed. Anyway, here is a solution with only simple angle chasing. Let $O_1$ be the center of $\alpha$. Clearly $ABEF$ is concyclic, all the points are on $\alpha$. $$\angle EAB=\angle EFB$$Now notice the quadrilateral $O_{1}APB$, it is concyclic as all the points are on $\omega$. We have $2\angle AEB=\angle AO_{1}B$ And $$\angle AO_{1}B=180 - \angle APB$$$$\angle APB=\angle EPB=180 - 2\angle PEB$$Now notice $\triangle EPB$, here $$\angle EBP=180-(180-2\angle PEB)-\angle EPB=\angle EPB$$So $\angle EAB=\angle EFB$ and $\angle AEB=\angle PEB=\angle EBP=\angle EBF$, and as they share the side $EB$, $\triangle AEB\cong \triangle FEB$ so $EF=AB$.

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SOLUTION 1 ( by circular inversion) Let O be the center of w1 and define OP ∩ AB as Q Consider a circular inversion with respect to w1 It is obvious that AB is the inverse of w2 and Q is the inverse of P with respect to w1 So <APO = < BPO =<ABO= <BAO and <ABF=AEF Since APOF is cyclic so <OBF = <EAO = <OEA = <OFB And,<BOF = <EOA so <APF + <AOB = <APF + <EOF = 180° So EOPF is cyclic and Q is the radical center of w1, w2, (EOPF) So E, P, F are collinear MOREOVER, POWw1(P)=PE.PA = PB.PF so QA.QE = QB.QF( inversion formula) Also POWw1(Q) =QE.QF = QA.QB Combining these results yields AQ = FQ and BQ = EQ So AQ + BQ = EQ + FQ So AB = EF (as desired)

SOLUTION 2 (by trigonometry) EOFP is cyclic and E,Q,F are collinear as in SOL 1 now (sin<EPO)/(sin<BPO) . (sin<PBA)/(sin<ABO) .(sin<OEF)/(sin<FEP) = 1 By coverse of trigonometric form of Cevas theorem, E , O , B are collinear BY extended law of sines, 2R = AB/(sin<AEB) = EF/(sin<EBF) <AEB = <EBF so AB = EF (as desired)

SOLUTION 1 ( by circular inversion) Let O be the center of w1 and define OP ∩ AB as Q Consider a circular inversion with respect to w1 It is obvious that AB is the inverse of w2 and Q is the inverse of P with respect to w1 So <APO = < BPO =<ABO= <BAO and <ABF=AEF Since APOF is cyclic so <OBF = <EAO = <OEA = <OFB And,<BOF = <EOA so <APF + <AOB = <APF + <EOF = 180° So EOPF is cyclic and Q is the radical center of w1, w2, (EOPF) So E, P, F are collinear MOREOVER, POWw1(P)=PE.PA = PB.PF so QA.QE = QB.QF( inversion formula) Also POWw1(Q) =QE.QF = QA.QB Combining these results yields AQ = FQ and BQ = EQ So AQ + BQ = EQ + FQ So AB = EF (as desired)SOLUTION 1 ( by circular inversion) Let O be the center of w1 and define OP ∩ AB as Q Consider a circular inversion with respect to w1 It is obvious that AB is the inverse of w2 and Q is the inverse of P with respect to w1 So <APO = < BPO =<ABO= <BAO and <ABF=AEF Since APOF is cyclic so <OBF = <EAO = <OEA = <OFB And,<BOF = <EOA so <APF + <AOB = <APF + <EOF = 180° So EOPF is cyclic and Q is the radical center of w1, w2, (EOPF) So E, P, F are collinear MOREOVER, POWw1(P)=PE.PA = PB.PF so QA.QE = QB.QF( inversion formula) Also POWw1(Q) =QE.QF = QA.QB Combining these results yields AQ = FQ and BQ = EQ So AQ + BQ = EQ + FQ So AB = EF (as desired)

SOLUTION 2 (by trigonometry) EOFP is cyclic and E,Q,F are collinear as in SOL 1 now (sin<EPO)/(sin<BPO) . (sin<PBA)/(sin<ABO) .(sin<OEF)/(sin<FEP) = 1 By coverse of trigonometric form of Cevas theorem, E , O , B are collinear BY extended law of sines, 2R = AB/(sin<AEB) = EF/(sin<EBF) <AEB = <EBF so AB = EF (as desired)