lasha.p wrote: we want to prove thet BAIK is cyclic,if and only if when CJDK is cyclic... it's <=> thet to prove that JA*JB=JI*JK <=> DI*IC=KI*IJ. FJ=z, IF=x and KI=y.. so EK=x+y. ABFE is cyclic,so JA*JB=z(2x+2y+z). <BEI=<<FAJ=<FDI so ECFD is cyclic and DI*IC=FI*IE=x(x+2y) so we want to prove that x(x+2y)=y(x+z) <=> z(2x+2y+z)=(z+x)(x+y+z) .... z(2x+2y+z)=(z+x)(x+y+z) <=> z(x+y+z) +z(x+y)= z(x+y+z)+x(x+y+z) <=> z(x+y)=x(x+y+z) <=> zy= x(x+y) and this is <=> x(x+2y)=y(x+z). lasha.p, if you repost this under your member name, I will delete this copy. Next time, please, use the <Reply to topic> button when posting a solution.

Since $\angle DAE \equiv \angle CBF$, $AEBF$ is cyclic, so $AJ \cdot JB = FJ \cdot JE$. Now, $I \in \odot ABK \Leftrightarrow \underline{IJ \cdot JK} = AJ \cdot JB = \underline{FJ \cdot JE}$ $\Leftrightarrow$ $IJ(\frac{IF+IE}{2}-IJ)=(IJ-IF)(IE-IJ)$ $\Leftrightarrow$ $IJ = \frac{2IF \cdot IE}{IF + IE}$ $\Leftrightarrow$ $IF \cdot IE = IJ \cdot \frac{IE+IF}{2}$ $\Leftrightarrow$ $IE \cdot IF = IJ \cdot IK$. Since $AEBF$ is cyclic, we get $\angle FAB \equiv \angle FEB \Rightarrow \angle IDF \equiv \angle FEC$ $\Leftrightarrow$ $FECD$ cyclic $\Leftrightarrow$ $IF \cdot IE = ID \cdot IC$. So, $IJ \cdot IK = IE \cdot IF$ $\Leftrightarrow$ $IJ \cdot IK = ID \cdot IC$ $\Leftrightarrow$ $K \in \odot CDJ$.

Computational solution: first notice that $FCED$ is cyclic. in deed $\angle DFE =\angle ABC=\angle DCE$ We have to prove that (with power of a point ): \[IJ.IK=JF.JE \Leftrightarrow IE.IF=IJ.IK\] $IJ.IK=JF.JE \Leftrightarrow IJ(\frac{FE}{2}-IF )=IE.IF \Leftrightarrow \frac{IJ.FE}{2}=IF.JE $ similarly $IE.IF=IJ.IK \Leftrightarrow \frac{IJ.FE}{2}=IF.JE $ so the equivalence is proved . QED

Inversive solution: (i think that K being the midpoint is not a necessary condition ) $ \angle DFE =\angle ABC=\angle DCE \Rightarrow FCED$ is cyclic. Consider $\psi$ the inversion centered at J and of ratio $\sqrt{JA.JB}$. and then $\psi(A)=B$ and $\psi(E)=F$ Let $\psi(C)=C'$ and $\psi(B)=B'$, and $\psi(I)=K'$ ($K'$ is the intersection of $(AIB)$ with $JC$) The image of $(FCED)$ must be a circle so $FC'ED'$ is cyclic. then the radical axis of circle $(FC'ED'), \ (DC'CD')$ and $(FCED)$ are concurent. since $FE\cap DC= I$, we get $I\in D'C'$ As a consequence $JDK'C$ is cyclic. Therefore : \[ABKI \ cyclic \Leftrightarrow K=K' \Leftrightarrow JDKC\ cyclic \] QED. In other words $(JDC)$ and $(AIB)$ intersect in segment $FE\blacksquare .$

$K \in \bigcirc CDJ \iff IJ \cdot IK=ID \cdot IC$ $\iff IJ \cdot IK = IF \cdot IE $ (As $EFDC$ is concyclic)$ \iff (I,J;F,E)= (-1)$(As $K$ is midpoint of $FE$)$\iff IJ \cdot JK=FJ \cdot JE \iff IJ \cdot JK=AJ \cdot JB$ (As $AEBF$ is concyclic.) $\iff I \in \bigcirc AKB$.

Legend-crush wrote: Inversive solution: (i think that K being the midpoint is not a necessary condition ) $ \angle DFE =\angle ABC=\angle DCE \Rightarrow FCED$ is cyclic. Consider $\psi$ the inversion centered at J and of ratio $\sqrt{JA.JB}$. and then $\psi(A)=B$ and $\psi(E)=F$ Let $\psi(C)=C'$ and $\psi(B)=B'$, and $\psi(I)=K'$ ($K'$ is the intersection of $(AIB)$ with $JC$) The image of $(FCED)$ must be a circle so $FC'ED'$ is cyclic. then the radical axis of circle $(FC'ED'), \ (DC'CD')$ and $(FCED)$ are concurent. since $FE\cap DC= I$, we get $I\in D'C'$ As a consequence $JDK'C$ is cyclic. Therefore : \[ABKI \ cyclic \Leftrightarrow K=K' \Leftrightarrow JDKC\ cyclic \]QED. In other words $(JDC)$ and $(AIB)$ intersect in segment $FE\blacksquare .$ The solution is wrong. circles (FC'ED'), (DC'CD'), (FCED) are duplicate

Please Who can draw figure. When I draw one quadrilateral is cyclic but another is not even convex.

My solution: From condition $FABE$ is cyclic or $JF\cdot JE=JA\cdot JB.(1).$ Also we know $180-\angle EFD=\angle EFA=180-\angle ABC=\angle BCD=180-\angle DCE\to DFCE $ is cyclic,or $DI\cdot IC=FI\cdot IE.(2).$ $ABKI$ is cyclic iff $$JI\cdot JK=JA\cdot JB=^{(1)}JF\cdot JE=(JK-FK)(JK+FK)=JK^2-FK^2\to JK\cdot IK=KF^2(3).$$Also $CKDJ$ is cyclic iff $$JI\cdot IK=DI\cdot IC=^{(2)}FI\cdot IE=(KF-IK)(KF+IK)=KF^2-IK^2\to KF^2=JI\cdot IK+IK^2=IK\cdot JK.(4).$$From $(3),(4)$ as desired.

*Flashback 2 years ago.* April wrote: Given trapezoid $ ABCD$ with parallel sides $ AB$ and $ CD$, assume that there exist points $ E$ on line $ BC$ outside segment $ BC$, and $ F$ inside segment $ AD$ such that $ \angle DAE = \angle CBF$. Denote by $ I$ the point of intersection of $ CD$ and $ EF$, and by $ J$ the point of intersection of $ AB$ and $ EF$. Let $ K$ be the midpoint of segment $ EF$, assume it does not lie on line $ AB$. Prove that $ I$ belongs to the circumcircle of $ ABK$ if and only if $ K$ belongs to the circumcircle of $ CDJ$. Proposed by Charles Leytem, Luxembourg Notice that $ABFE$ is a cyclic quadrilateral. Consider the following Lemma. $CDFE$ is cyclic. (Proof) Apply the converse of Reim's Theorem to $ABEF$ and lines $\overline{BE}$ and $\overline{AF}$. This works well since $\overline{AB} \parallel \overline{CD}$. $\blacksquare$ By power of point, $$K \in (CDJ) \iff IJ \cdot IK =IC \cdot ID=IF \cdot IE \iff (FE, IJ)=-1,$$and $$I \in (ABK) \iff JI \cdot JK=JA\cdot JB=JF\cdot JE \iff (FE, IJ)=-1.$$Hence, both conditions are equivalent. $\blacksquare$

anantmudgal09 wrote: *Flashback 2 years ago.* April wrote: Given trapezoid $ ABCD$ with parallel sides $ AB$ and $ CD$, assume that there exist points $ E$ on line $ BC$ outside segment $ BC$, and $ F$ inside segment $ AD$ such that $ \angle DAE = \angle CBF$. Denote by $ I$ the point of intersection of $ CD$ and $ EF$, and by $ J$ the point of intersection of $ AB$ and $ EF$. Let $ K$ be the midpoint of segment $ EF$, assume it does not lie on line $ AB$. Prove that $ I$ belongs to the circumcircle of $ ABK$ if and only if $ K$ belongs to the circumcircle of $ CDJ$. Proposed by Charles Leytem, Luxembourg Notice that $ABFE$ is a cyclic quadrilateral. Consider the following Lemma. $CDFE$ is cyclic. (Proof) Apply the converse of Reim's Theorem to $ABEF$ and lines $\overline{BE}$ and $\overline{AF}$. This works well since $\overline{AB} \parallel \overline{CD}$. $\blacksquare$ By power of point, $$K \in (CDJ) \iff IJ \cdot IK =IC \cdot ID=IF \cdot IE \iff (FE, IJ)=-1,$$and $$I \in (ABK) \iff JI \cdot JK=JA\cdot JB=JF\cdot JE \iff (FE, IJ)=-1.$$Hence, both conditions are equivalent. $\blacksquare$ What kind of Reim's theorem did you use here?

Here's a PoP bashing solution. Note the condition $\angle DAE = \angle CAF$ implies the four points $F, A, B, E$ are concyclic. Also $\angle FDC = \angle DAJ = 180 - \angle FAB = \angle FEB = \angle FEC$, hence $\omega_{FDEC}$ exists too. By PoP, we get $$ID \cdot IC = IF \cdot IE = (FK - IK)(EK+IK) = (EK-IK)(EK+IK) = EK^2 - IK^2 (\star)$$ For the first part, i.e $I \in \omega_{ABK} \Rightarrow K \in \omega_{CDJ}$. Note that $ \text{Pow}_J(\omega_{ABEF}) = JF \cdot JE = (JK + EK)(JK - EK) = JK^2 - EK^2 $ and $\text{Pow}_J(\omega_{ABK}) = JI \cdot JK = (JK - KI) \cdot JK = JK^2 - JK \cdot KI = JK^2 - ((JI + IK) \cdot KI) = JK^2 - (JI \cdot KI + KI^2)$. But $J$ lies in the radical axis of $\omega_{ABEF}$ and $\omega_{ABK}$. Hence, the powers are equal, which gives $EK^2 = JI \cdot KI + KI^2$, or $ IJ \cdot IK= EK^2 - KI^2 \overset{\star}{=} ID \cdot IC $, which implies $J, K, D, C$ are concyclic, as desired. For the second part (i.e the converse), assume $J, K, D, C$ are concyclic and we need to prove that $A, B, K, I$ are concyclic. Since $C,J, D, K$ are concylic, $IJ \cdot IK = \text{Pow}_I(\omega_{CDJK}) = ID \cdot IC \overset{\star}{=} EK^2 - IK^2$. Also $IJ \cdot IK = (JK - IK)IK = JK \cdot IK - IK^2$. Combining these two, we get $EK^2 = JK \cdot IK (\spadesuit)$. So $JA \cdot JB = JF \cdot JE = JK^2 - EK^2 \overset{\spadesuit}{=} JK^2 - JK \cdot IK = JK(JK-IK) = JK \cdot IJ$, hence $A, B, K, I$ are concyclic, as desired.

It is the part of IMOSL 2004 G8. Lemma: Given a cyclic quadrilateral $ABCD$, let $M$ be the midpoint of the side $CD$. Let $BC\cap DA=F$ and $BD\cap CA=E$. Then $(ABM)\cap CD\in EF$ . Proof (by grobber): grobber wrote: Let $P$ be the second point of intersection between $CD$ and the circle $(ABM)$, and let $G=AB\cap CD$. A very simple computation, based on the fact that $GD\cdot GC=GA\cdot GB=GM\cdot GP$ and $M$ is the midpoint of $CD$ will show that $P$ is, in fact, the harmonic conjugate of $C,D$ art $G$, so it belongs to $EF$. So in both cases $(F,E;I,J)=-1$. So we are done.

My solution: We start off with the following well known lemma. Lemma Let $P,R,Q,S$ be points on a line in that order. Let $T$ be the midpoint of $RS$. Suppose that $PT \cdot PQ=PR \cdot PS$. Then $(P,Q;R,S)=-1$. PROOF: Notice that $PT \cdot PQ=PR \cdot PS=(PT-TR)(PT+TS)=(PT-TR)(PT+TR)=PT^2-TR^2$ $\Rightarrow PT(PT-PQ)=TR^2 \Rightarrow TP \cdot TQ=TR^2$. This means that $P$ and $Q$ are inverses wrt $\odot (RS)$, giving that $(P,Q;R,S)=-1$. $\blacksquare$ Return to the problem at hand. Note that, from the given conditions, $ABEF$ is cyclic. This also gives that $\measuredangle CDF=\measuredangle BAF=\measuredangle CEF \Rightarrow CEDF$ is cyclic. Now, $ABIK$ is cyclic $\Leftrightarrow JK \cdot JI=JA \cdot JB=JE \cdot JF \Leftrightarrow (J,I;E,F)=-1$, where the last equality follows from our Lemma. Similarly, $CDJK$ is cyclic $\Leftrightarrow IK \cdot IJ=IC \cdot ID=IE \cdot IF \Leftrightarrow (I,J;E,F)=-1$, where the last equality follows from our Lemma. Thus, The two given statements are equivalent to each other.

How is this even a G2? April wrote: Given trapezoid $ ABCD$ with parallel sides $ AB$ and $ CD$, assume that there exist points $ E$ on line $ BC$ outside segment $ BC$, and $ F$ inside segment $ AD$ such that $ \angle DAE = \angle CBF$. Denote by $ I$ the point of intersection of $ CD$ and $ EF$, and by $ J$ the point of intersection of $ AB$ and $ EF$. Let $ K$ be the midpoint of segment $ EF$, assume it does not lie on line $ AB$. Prove that $ I$ belongs to the circumcircle of $ ABK$ if and only if $ K$ belongs to the circumcircle of $ CDJ$. Proposed by Charles Leytem, Luxembourg By looking at the angles, we find that $ABEF$ is cyclic hence so is $DECF.$ Thus, \begin{align*} I \in \odot(ABK) &\Leftrightarrow JI \cdot JK=JA \cdot JB= JE \cdot JF &\Leftrightarrow (J,I;F,E)=-1 \\ K \in \odot(CDJ) &\Leftrightarrow IK \cdot IJ=IC \cdot ID=IF \cdot IE &\Leftrightarrow (I,J;F,E)=-1 \end{align*}And so we are done. $\square$

Note that $(DFEC)$ and $(AEFD)$. So, $$IJ\cdot JK=AJ\cdot JB\iff IJ\cdot JK=FJ\cdot JE=(FK)^2-JK^2\iff IK\cdot JK=JK^2$$$$\iff IK^2-JK^2=IK^2-IK\cdot JK\iff IF\cdot IE=IK\cdot IJ\iff IJ\cdot IK=ID\cdot IC$$as desired.

Note that $(ABFE)$ is cyclic. Now, here is a much more natural restatement of the problem: Quote: Points $A,B,F,E$ lie on a circle. Let $K$ be the midpoint of $EF$ and $J=EF\cap AB$. Let $\ell$ be the line through $I$ parallel to $AB$. Let $C=\ell\cap BE, D=\ell\cap AF$. Prove $CKDJ$ cyclic. Note $\angle DFE=180-\angle AFE=\angle ABE = \angle DCE$, so $DCEF$ cyclic. We want to show $IK\cdot IJ=ID\cdot IC = IF\cdot IE$. Since $(EF;K\infty)=-1$, inverting at $X$ swapping $E,F$ gives us that we want to show $(FE;JI)=-1$. Since $ABFE$ cyclic, $JF\cdot JE = JA\cdot JB = JI\cdot JK$. Again, since $(EF;K\infty)=-1$, inverting at $J$ swapping $E,F$ gives $(FE;IJ)=-1$, so $(FE;JI)=-1$, and we are done.

Can somone attach a complete diagram please? Spent half an hour using Geoegebra and couldn't even get the diagram.

@above diagram. You may also see the video.

By definition $ABEF$ is cyclic. Furthermore $\angle AFE = \angle BCD = 180^{\circ} - \angle ABC$ hence $DECF$ is cyclic. Now we are ready to PoP:\[I \in (ABK) \iff JI \cdot JK = JA \cdot JB = JF \cdot JE\]\[K \in (CDJ) \iff IF \cdot IE = ID \cdot IC = IK \cdot IJ\]and in fact both of these are the same condition as $(I, J; F, E) = (J, I; F, E) = -1$ and we are done. $\blacksquare$

Note that $AEBF$ is cyclic by the given angle conditions. Consider the negative inversion at $K$ mapping $(AEBF)$ to itself. Note that $I \in (ABK) \iff I^* = E^*F^* \cap A^*B^*$ . By butterfly theorem, $E^*F^* \cap A^*B^*$ is the reflection of $J$ over $K$, so $I \in (ABK)$ if and only if $I$ is the inverse of $J$ under a (positive) inversion at $K$ with radius $KF$. The same argument shows that $J \in (CDK)$ if and only if it is the inverse of $I$ under this inversion.

First, notice that $$\angle FAE=\angle DAE=\angle CBF = 180^{\circ}-\angle EBF,$$so $AEBF$ is cyclic. In addition, $$\angle CDF = \angle CDA=180^{\circ}-\angle BAD =180^{\circ}-\angle BAF = 180^{\circ}-\angle BEF=180^{\circ}-\angle CEF,$$so $DFEC$ is cyclic. Notice that $IF\cdot IE = ID\cdot IC$ by Power of a Point, and that $IJ\cdot IK = ID\cdot IC$ iff $DJKC$ is cyclic by Power of a Point and it's converse. Thus $IF\cdot IE = IJ\cdot IK$ is equivalent to $DJKC$ being cyclic. Also, $AJ\cdot JB = JF\cdot JE$ by Power of a Point, and $AKBI$ is cyclic iff $FJ\cdot JE=JI\cdot JK$ by Power of a Point and it's converse, so $AKBI$ cyclic is equivalent to $IJ\cdot JK = FJ\cdot JE$. Thus, it remains to prove the following claim: Claim: $IF\cdot IE= IJ\cdot IK$ is equivalent to $IJ\cdot JK=FJ\cdot JE$. Proof. Use coordinates so that the lines $AB$ and $CD$ coincide with $y=0$ and $y=-1$ respectively. Thus $I$ and $J$ have y-coordinates $-1$ and $0$, respectively. Let the y-coordinates of $F$ and $E$ be equal to $a$ and $b$, respectively. Then the y-coordinate of $K$ is equal to $\frac{a+b}{2}$. Notice that the distance between two points on line $EF$ is proportional to the difference in their y-coordinates. Thus $IF\cdot IE=IJ\cdot IK$ is equivalent to $$(a+1)\cdot (b+1)=1\cdot \left(\frac{a+b}{2}+1\right)\Rightarrow ab=-\frac{a+b}{2}.$$However, $IJ\cdot JK=FJ\cdot JE$ is equivalent to $$1\cdot \left(\frac{a+b}{2}+1\right) = -a\cdot b,$$which reduces to the same condition, so the claim is proved. $\blacksquare$ As the claim was the last step we needed to prove in our chain of equivalences, we are done.

The angle condition is equivalent to $AEBF$ cyclic. Now $$I \in (ABK) \iff JA \cdot JB = JK \cdot JI \iff JE \cdot JF = JK \cdot JI \iff (EF;JI)=-1.$$$DFCE$ is cyclic by angle chasing, so $$IJ \cdot IK = ID \cdot IC \iff IJ \cdot IK = IF \cdot IE \iff (EF;JI)=-1.$$These are clearly equivalent.

The angle condition is equivalent to $AEBF$ being cyclic, and $CDFE$ is cyclic as $$\angle FEC=\angle FAB=180-\angle CDF.$$By PoP, \begin{align*}I\in(ABK)&\iff IJ\cdot JK=AJ\cdot JB=FJ\cdot JE\\&\iff(IJ;EF)=-1\\&\iff IJ\cdot IK=IF\cdot IE=ID\cdot IC\\&\iff K\in(CDJ)\end{align*}$\square$

Clearly $F\in \odot (ABE)$ and by Reim $F\in \odot (CDE).$ $$I\in \odot (ABK)\iff |JI|\cdot |JK|=|JA|\cdot |JB|=|JE|\cdot |JF|\iff (EFIJ)=-1\iff$$$$\iff |IJ|\cdot |IK|=|IE|\cdot |IF|=|IC|\cdot |ID|\iff K\in \odot (CDJ)\text{ } \blacksquare$$

Note that $ABEF,CFDE$ cyclic. Consider the statement $(*):\frac{JF-IF}{IF}=\frac{IF}{IK}.$ Note that \[JF\cdot IK-IF^2=FJ\cdot FJ-IF\cdot IJ=FJ(JE-IK)-IJ^2=JF\cdot JE+IK\cdot IF-JI\cdot JK\]which when $JI\cdot JK=JF\cdot JE$ is equal to $IK\cdot IF$ which means that $I\in(ABK)\iff (*).$ On the other hand, \[IF^2-IK\cdot FJ=IF\cdot FK-IK\cdot IJ\]which when $IE\cdot IF=IJ\cdot IK$ is equal to $-IF\cdot IK$ which means $K\in(CDJ)\iff (*).$ Thus, we are done.

the angle condition implies that ABEF is cyclic , and since AB // CD , DECF is also cyclic . By power of a point : $$ABIK \quad \text{cyclic } \iff JF.JE=JA.JB =JK.JI $$ OR $$JF.JE=JK.JI \iff (FE;IJ) = -1 \iff IK.IJ= IE.IF=ID.IC $$Which is equivalent to DECF to be cyclic .

Note that $ABEF$ is cyclic from the given angle condition. Then by Reim's theorem, $CDEF$ must also be cyclic. From PoP, $K$ lies on $ABI$ if and only if \[ JI \cdot JK = JA \cdot JB = JE \cdot JF \]which is equivalent to $(JI;EF) = -1$ (from a lemma regarding harmonic bundles). Additionally, note from PoP that $K$ lies on $CDJ$ if and only if \[ IE \cdot IF = IC \cdot ID = IK \cdot IJ \]But $IE \cdot IF = KE^2 - KI^2$ from PoP, so the condition is true if and only if $KE^2 = KI \cdot KJ$, which is true if and only if $(JI;EF) = -1$. $\blacksquare$

Note that the angle condition rewrites as points $A$, $E$, $B$ and $F$ being concyclic. By Reims, we further have $D$, $F$, $E$ and $C$ concyclic. Then, by PoP and Mac-Laurin, both circles in the statement exist if and only if $I$, $J$, $F$ and $E$ are harmonic. $\square$

The angle condition gives $AEBF,CEFD$ cyclic. $CDJK$ cyclic is equivalent to $IJ\cdot IK=IC\cdot ID=IE\cdot IF=IK^2-KE^2.$ $AKBI$ cyclic is equivalent to $JK\cdot JI=JA\cdot JB=JE\cdot JF=KE^2-JK^2.$ Adding these gives the identity so we are done.

Notice $ABEF$ cyclic. By PoP we get that $JK \cdot JI = JA \cdot JB = JF \cdot JE \implies (J, I; F, E) = -1$. Also we get that $\angle DAB = 180^\circ - \angle FEB = \angle ADC$ so $DFCE$ is cyclic. PoP gives us $IF \cdot IE = IC \cdot ID$ which by $-1 = (J, I; F, E)$ is equivalent to $IK \cdot IJ$, giving $CKDJ$ cyclic as desired.

means $-1=(EF; IJ)$. Also note that by an analagous reason $CKDJ$ is cyclic iff $-1=(EF;IJ)$ thus both cyclicities imply each other.

Since $\angle DAE = \angle CBF$ we have ABEF cyclic. Since $AB \parallel CD$ we have CDEF cyclic by Reim's. Then, \[ I \in (ABK) \iff JB \cdot JA = JI \cdot JK \]But since we also have $JB \cdot JA = JF \cdot JE$ by PoP, this is equivalent to $(E, F;I, J) = -1$. Using a similar approach we have that $J \in (CDK)$ is also equivalent to $(E, F;I, J) = -1$, done.

We can prove these two directions separately. First, however, let us make the following claim. *** Claim 1. If $(ABK)$ intersects $EF$ at a point $I'$ such that $I'\neq K$, then it holds that $(EF;I'J)=-1$. Proof. This is true by part of the configuration seen in ISL 2004/G8. *** We now will prove our directions. *** 1. $I\in (ABK)\implies K\in (CDJ)$. Proof. Since $I\in (ABK)$, we must have that $I=I'$. Now, note that since \[\angle DAE=\angle CBF,\]we must have that $ABEF$ is cyclic. However, by Reim's Theorem, since $AB\parallel CD$, we also have that $CEDF$ is cyclic. Now, by Power of a Point, we have that \[K\in (CDJ) \iff IC*ID=IK*IJ,\]however, since $CEDF$ is cyclic, we have that \[IC*ID=IE*IF,\]so our problem becomes proving that $IK*IJ=IE*IF$. Knowing that $K$ is the midpoint of $EF$ and $(EF;IJ)=-1$ (by Claim 1). WLOG letting $IE=a$ and $IF=1$, we can set up a "number line" with directed lengths and label the (directed) lengths as follows: [asy][asy] draw((0,0)--(12,0)); dot((0,0)); dot((6,0)); dot((8,0)); dot((4,0)); dot((12,0)); label("$E$", (0,0), S); label("$I$", (6,0), S); label("$F$",(8,0),S); label("$J$",(12,0),S); label("$K$", (4,0),S); label("$\frac{a+1}{2}$", (2,0), N, red); label("$\frac{a-1}{2}$", (5,0), N, red); label("$1$", (7,0), N, red); label("$\frac{a+1}{a-1}$", (10,0), N, red); [/asy][/asy] Then, through simple computation, we get that \[IE*JF=a*1=\frac{a-1}{2}*\frac{2a}{a-1}=IJ*IK,\]as desired. This means that $I\in (ABK)\implies K\in (CDJ)$, proving our first direction. *** We will now prove the other direction. *** 2. $K\in (CDJ)\implies I\in (ABK)$. Proof. Let $CD$ intersect $EF$ at the point $I''$. We aim to show that $I''=I'$, which shows that $I\in (ABK)$, as desired. Since $(EF;I'J)=-1$, note that by the uniqueness of the harmonic conjugate, it then suffices to show that $(EF;I''J)=-1$ also. Since $CKDJ$ is cyclic, by Power of a Point, we have that \[I''C*I''D=I''K*I''J.\]Again, by Reim's, note that $CEDF$ is cyclic. This gives us that \[I''C*I''D=I''K*I''J=I''E*I''F \implies \frac{I''E}{I''J}=\frac{I''K}{I''F}.\]Now, knowing this, and that $K$ is the midpoint of $EF$, we would like to prove that $(EF;I''J)=-1$. We can again WLOG let $I''K=a'$ and $I''F=1$ and set up a "number line" with directed lengths and label the (directed) lengths as follows: [asy][asy] draw((0,0)--(12,0)); dot((0,0)); dot((6,0)); dot((8,0)); dot((4,0)); dot((12,0)); label("$E$", (0,0), S); label("$I$", (6,0), S); label("$F$",(8,0),S); label("$J$",(12,0),S); label("$K$", (4,0),S); label("$a'+1$", (2,0), N, red); label("$a'$", (5,0), N, red); label("$1$", (7,0), N, red); label("$\frac{a'+1}{a'}$", (10,0), N, red); [/asy][/asy] Then, again through computation, we get that \[I''E*JF=\frac{a'+1}{a'}*(2a'+1)=1*\frac{(a'+1)(2a'+1)}{a'}=I''F*JE,\]which means that $\frac{I''E}{I''F}=\frac{JE}{JF}$, or $(EF;I''J)=-1$, as desired. This means that $I''=I'$, meaning that $I$ does indeed lie on $(ABK)$, given that $K\in (CDJ)$. This proves our second direction. *** Now, since we have proved both directions, we have that $I\in (ABK)\iff K\in (CDJ)$, as we wished to prove, completing our proof. $\blacksquare$

The angle condition and Reim's gives $ABEF$ and $CDEF$ cyclic. We now notice \begin{align*} &I \in (ABK) \iff JI \cdot JK = JA \cdot JB \iff JI \cdot JK = JE \cdot JF \\ &K \in (CDJ) \iff IJ \cdot IK = IC \cdot ID \iff IJ \cdot IK = IE \cdot IF, \end{align*} which are both equivalent to $(IJ;EF) = -1$. $\blacksquare$