Let $A'$ be the reflection of $A$ over $BD$. We show the circles are tangent at $A'$. Indeed, $BEKA'$ is cyclic because$$\measuredangle BA'K = -\measuredangle BAK = \measuredangle BEA = \measuredangle BEK$$and similarly $DFLA'$ is cyclic. Hence by reflection across $BD$ it suffices to show $(ABK)$ and $(ADL)$ are tangent at $A$: let the tangents at $A$ to these circles be $\ell_1, \ell_2$ respectively. Indeed, note \begin{align*} \measuredangle (\overline{AB}, \ell_1)+ \measuredangle (\ell_2, \overline{AD}) &= \measuredangle BKA + \measuredangle ALD \\ &= \measuredangle LKA + \measuredangle ALK \\ &= \measuredangle LAK = \measuredangle (\overline{CD}, \overline{CB})\\ &= \measuredangle (\overline{AB}, \overline{AD}) \end{align*}hence $\ell_1=\ell_2$.

Cool problem. $\textbf{Lemma 1:}$ The circumcircles of $\triangle ADL$ and $\triangle ABK$ are tangent at $A$ $\textbf{Proof)}$ Notice that $$\angle LAK = 180^{\circ}-\angle CDB = 180^{\circ} - \angle DAB$$We consequently have that $$\angle DKA + \angle ADK = 180^{\circ} - \angle DAK = \angle LAB$$Now, if we select a point $T$ inside $\triangle DAK$ such that $AT$ is tangent to the circumcircle of $\triangle DAL$, we have that $$ \angle KAB = \angle DKA = \angle LAB - \angle KDA = \angle BAT + \angle LAT - \angle DLA = \angle BAT$$meaning that $AT$ is also tangent to the circumcircle of $\triangle BAK$. $\square$ $\textbf{Lemma 2:}$ The circumcircles of $\triangle DFL$ and $\triangle BKE$ are tangent $\textbf{Proof)}$ Notice that if we let $A'$ be the reflection of $A$ across $BD$, then $$\angle DA'L = \angle DAL = \angle DFL$$meaning that $A'$ lies on the circumcircle of $\triangle DFL$. Analagously, $A'$ also lies on the circumcircle of $\triangle BKE$. Moreover, notice that the circumcircles of $\triangle A'LD$ and $\triangle A'BK$ are tangent by $\textbf{Lemma 1}$ meaning indeed that the circumcircles of $\triangle BEK$ and $\triangle DFL$ are tangent to each other. $\blacksquare$

Solved with dantaxyz.

Let $A'$ be the reflection of $A$ over $\overline{BD}$. We claim $A'$ is the desired tangency point. First, we show it lies on each of the circles. Let $A_1$ and $E_1$ be the feet from $A$ to $\overline{BD}$ and $\overline{BC}$ respectively. Noticing that $AE_1BA_1$ is cyclic, note \begin{align*} \angle KEA' = \angle AEA' &= \angle AE_1A_1 = \angle ABD \\ &= 180^\circ-\angle KBA = 180^\circ-\angle KBA', \end{align*}so $KBA'E$ is cyclic. Hence $A' \in (BEK), (DFL)$. To show the tangency at $A'$, it suffices to show \[ \tfrac12 \widehat{BA'} + \tfrac12 \widehat{DA'} = \angle BA'D. \]We have $\tfrac12 \widehat{BA'}=\angle BKA'=\angle BKA=90^\circ-\angle DBC$, and similarly $\tfrac12 \widehat{DA'} = 90^\circ-\angle BDC$, so their sum is $(90^\circ-\angle DBC)+(90^\circ-\angle BDC)=\angle BCD = \angle BAD$, since $\angle A=\angle C$, and we are done.

[asy][asy] size(8cm); defaultpen(fontsize(10pt)); pair A = dir(110), B = dir(195), D = dir(345), C = 2*foot(dir(80),B,D)-dir(80), P = foot(A,B,D), A1 = 2P-A, E = 2*foot(A,B,C)-A, F = 2*foot(A,C,D)-A, K = extension(B,D,A,E), L = extension(A,F,B,D), Ob = circumcenter(A1,B,E), Od = circumcenter(A1,D,F), B1 = extension(B,C,A,E), D1 = extension(C,D,A,F); draw(B--B1^^D--D1, purple+dotted); draw(A--B--C--D--A, purple); draw(B--A1--D, purple); draw(K--L^^E--A--F^^A1--A, heavycyan); draw(circumcircle(B,C,D), heavygreen); draw(circumcircle(B,K,E)^^circumcircle(D,L,F), orange); draw(Ob--Od, magenta); dot("$A$", A, dir(90)); dot("$B$", B, dir(225)); dot("$C$", C, dir(270)); dot("$D$", D, dir(315)); dot("$A'$", A1, dir(255)); dot("$P$", P, dir(45)); dot("$E$", E, dir(150)); dot("$F$", F, dir(45)); dot("$B_1$", B1, dir(135)); dot("$D_1$", D1, dir(60)); dot("$K$", K, dir(135)); dot("$L$", L, dir(60)); dot("$O_B$", Ob, dir(165)); dot("$O_D$", Od, dir(345)); [/asy][/asy] Let $B_1$, $D_1$ be the midpoints of $\overline{AE}$, $\overline{AF}$ respectively. Let $P$ be the foot from $A$ to $\overline{BC}$ and $A'$ be the reflection of $A$ over $P$. I claim that $A'$ is the desired tangency point. Claim: $A’BKE$, $A’DLF$ are cyclic. Proof. Since $\overline{B_1P}$ is the $A$-midline in $\triangle AEA'$, we have $$\measuredangle KEA' = \measuredangle AEA' = \measuredangle AB_1P = \measuredangle ABP = \measuredangle PBA' = \measuredangle KBA',$$so $A’BKE$ is cyclic, and the other follows analogously. $\square$ To finish, let $O_B$ be the center of $(A'BKE)$ and $O_D$ be the center of $(A'DLF)$. Note that $B$ is the center of $(AEA')$, so $$\angle O_BA'B = \frac 12 \angle A'BE = \angle A'AE.$$Thus \begin{align*}\angle O_BA'B + \angle BA'D + \angle DA'O_D &= \angle A'AE + \angle BCD + \angle FAA' \\ &= \angle DBC + \angle BCD + \angle CDB \\ &= 180^\circ,\end{align*}which means $O_B$, $A'$, $O_D$ are collinear, and we are done.

The key observation is that the point of tangency is $A',$ the reflection of $A$ over $BD.$ Observe that $\measuredangle DFL = \measuredangle LAD=\measuredangle DA'L,$ so $A'\in (LDF)$ and similarly $A'\in (KBE).$ Now suppose that line $\ell$ is tangent to $(LDF)$ at $A'.$ Then we obviously have $\measuredangle (LA',\ell) = \measuredangle LDA' = \measuredangle ADB,$ but on the other hand we find that \begin{align*} \measuredangle LA'K &= \measuredangle KAL \\ &= \measuredangle BAD - \measuredangle BAH - \measuredangle IAD \\ &= \measuredangle BAD - (90^\circ - \measuredangle HBA) - (90^\circ - \measuredangle ADI) \\ &= \measuredangle BAD + (360^\circ - 2\measuredangle BAD) - 180^\circ \\ &= \measuredangle DBA + \measuredangle ADB, \end{align*}so in fact $\measuredangle (\ell, KA') = \measuredangle DBA = \measuredangle A'BK,$ meaning that $\ell$ is tangent to $(KBE)$ at $A'$ as well. $\blacksquare$

Really nice problem, though I drammatically missed it (and hence the IMO) after guessing the tangency point, more here (see Day 2).

Let $T$ be reflection of $A$ wrt $BD$. Then $\measuredangle DTL=\measuredangle FAD=\measuredangle DFL\implies T\in \odot (DFL)$. Analogously $T\in \odot (BEK)$. Finally $\measuredangle BTD=\measuredangle BCD=\measuredangle KAL=\measuredangle DLA+\measuredangle AKB=\measuredangle TLD+\measuredangle BKT$, hence $T$ is desired tangency point.

If $A'$ is the reflection of $A$ wrt $BD$ then $\angle BA'K=\angle BAK=\angle BEK \implies A' \in (KBE)$. Similarly $A' \in (DLF)$. $\angle A'KB+\angle A'DL=180-(180-\angle BCD)=\angle BA'D$ which implies $A'$ is the desired tangency point.

turns out having a logic fault and assuming ABCD is a kite was actuallly pretty useful Denote $A'$ as the reflection of $A$ over $BD$. The claim is that $A'$ is the desired tangency point. First, we show that $A'$ lies on both circles. This follows from the fact that $B$ is the circumcenter of $AEA'$ --- assume this triangle is acute (the obtuse case follows similarly): \[\angle EBA'=2\angle A'AE=\angle KAA'+\angle KA'A=\angle EKA'\]which impllies the concylicity. $A'$ lying on $(DFL)$ follows similarly. Note that in order to show the two circles are tangent, it suffices to show $\angle A'EB+\angle A'FD=\angle BA'D$ (consider constructing the tangents at $A'$ for both circles). This is, again, angle chase using the two circumcenters $B$ and $D$: \[\angle A'EB+\angle A'FD=90^{\circ}-\angle EAA'+90^{\circ}-\angle FAA'=180^{\circ}-\angle EAF\]\[=90^{\circ}-\angle CAE+90^{\circ}-\angle CAF=\angle BCA+\angle DCA=\angle BCD=\angle BAD=\angle BA'D\]as desired.

Let $T$ be the reflection of $A$ over $\overline{BD}$. We claim that this is the desired tangency point. Observe that \[ \measuredangle DTL = -\measuredangle DAL = \measuredangle DFL, \]and hence $T$ lies on $(DFL)$. Similarly, $T$ also lies on $(BEK)$. Now reflect these two circles over $\overline{BD}$. It suffices to show that $(BKA)$ and $(DAL)$ are tangent to each other at $A$. Just notice that \[ \measuredangle ABK + \measuredangle DLA = \measuredangle BAL = \measuredangle BAK + \measuredangle KAL = \measuredangle BAK + \measuredangle BCD = \measuredangle BAK + \measuredangle DAB = \measuredangle DAK, \]and we are done.

Define $A'$ as the reflection of $A$ over $BD$. Observe that \[\angle KEA' = \angle ABP = 180 - \angle KBA'\]so $(EKBA')$ is cyclic. Similarly, $(FLDA')$ is cyclic. Now, if the centers of $(EKB)$ and $(FDL)$ were $O_{1}$ and $O_{2}$ respectively, then \[\angle O_{1}A'A + \angle O_{2}A'A = \angle O_{1}A'B + \angle O_{2}A'D + \angle BA'D = \angle CBD + \angle BDC + \angle BCD = 180\]Therefore, $O_{1}, A', O_{2}$ are collinear, so $(BKE)$ and $(DLF)$ are tangent at $A'$.

ok apparently this was generated by the same AI as G6 lmao

How do you know that?

Kagebaka wrote: ok apparently this was generated by the same AI as G6 lmao Originally, the software generated this figure (and text). After solving it, I slightly generalized it (and then the PSC reformulated it to the final statement)

Let reflection of $A$ over $BD$ be $A'$. Then $\measuredangle BEK = \measuredangle BEA = \measuredangle EAB = \measuredangle KAB = \measuredangle BA'K$ meaning $A'\in (BEK)$. Analogously, $A'\in (DFL)$. So by reflection over $BD$, it is enough to show $(ADL)$ and $(ABK)$ are tangent at $A$. By inversion at $A$, it would suffice to show $\angle ALD + \angle ADB + \angle ABD + \angle AKB=180^\circ$. Equivalently, we need to check $\angle ALD + \angle AKB = \angle BAD$. Let $AK\cap BC = S, AL\cap CD = T$. Then \[\angle ALD +\angle AKB = \angle TLD + \angle SKB = 90^\circ - \angle TDL + 90^\circ - \angle SBK = 90^\circ - \angle CDB + 90^\circ - \angle CBD = \angle BCD,\]done.

This was so nice. It's not that easy to prove tangency of two circles with about nothing to do with each other, and so we try to find the point of tangency first. Let $G$ be the reflection of $A$ across $BD.$ Now, we prove the point is on both circles. Note that $\angle KGB=\angle KAB=\angle KEB$ so $KEGB$ is cyclic. Similarly, $LFGD$ is cyclic. Now, we need to show that a line passing through the point of tangency is tangent to both circles, which reduces to a simple angle condition. Let $X$ be midpoint of $AE$, $Y$ midpoint of $AF.$ Note that $AXCY$ is cyclic. We have $\angle BGD=\angle BCD=180^\circ-\angle XAY=180^\circ-\angle KGL=\angle BKG+\angle DLG.$ Thus, there exists a line through $G$ tangent to both circles, so the two circles are tangent as desired.

The original post is missing two degree signs and one angle sign (at $\angle ABC<90^\circ$ and $\angle CDA<90^\circ$). Can anyone fix that?

Funny problem! Let $A'$ be the reflection of $A$ in $BC$. Note that $\angle KEB = \angle KAB = \angle KA'B$, so $KEA'B$ (and similarly $FLDA'$) are cyclic quadrilaterals. Thus, to prove the original problem, it suffices to show that $(AKB)$ and $(ADL)$ are tangent. Note that $\angle AKB = 90^{\circ} - \angle DBC$ and $\angle ALD = 90^{\circ} - \angle BDC$. Thus $\angle AKB + \angle ALD = \angle BAD$, so the circumcircles of triangles $AKB$ and $ALD$ are tangent. We are done.

Posting solution for storage even though it is similar to others on this thread Let $A'$ be the reflection of $A$ over $BD$. Claim 1: $BKA'E$ is cyclic. Proof: Notice that $\measuredangle BEK=\measuredangle BEA = \measuredangle BA'K=-\measuredangle BAK $, which implies the result. Similarly, quadrilateral $DFLA'$ is cyclic. Now, we can note that the problem is equivalent to proving that the circumcircle of $ABK$ and the circumcircle of $ADL$ are tangent at point $A$. Claim 2: $(ABK)$ is tangent to $(ADL)$ at $A$. Proof: Let the tangent to $(ABK)$ at $A$ be $l_{ABK}$ and the tangent to $(ADL)$ at $A$ be $l_{ADL}$. Let $K'$ and $L'$ be points on these lines, respectively. It follows that $\measuredangle BAK'+\measuredangle DAL'=\measuredangle BKA+\measuredangle ALD=\measuredangle AKL+\measuredangle ALK=\measuredangle LAK$. Because $AXCY$ is cyclic, we then have that $\measuredangle LAK=\measuredangle BCD=\measuredangle BAD$. This implies that $l_{ABK}\equiv l_{ALD}$, hence finishing the problem.

Let $X$ be the reflection point of $A$ wrt $BD, G$ be a point on $AE$ such that $GX$ is tangent to $(LDX)$ We have $\angle LXD=\angle LAD=\angle LFD\Rightarrow LDXF$ is cyclic, similarly $KEBX$ is cyclic Also $\angle ALK+\angle AKL=(\angle ADB-\angle LAD) + (\angle ABD-\angle KAB)=(\angle ADB+\angle ABD) - \angle LAD -\angle KAB$ $=(180^{\circ}-\angle DAB)-(\angle ADC-90^{\circ})-(\angle ABC-90^{\circ})=(360^{\circ}-\angle ADC-\angle ABC) -\angle DAB=2\angle DAB-\angle DAB=\angle DAB$ $\Rightarrow \angle XLD + \angle AKL=\angle DXB=\angle GXD +\angle GXB,$ but $\angle XLD=\angle GXD,$ as $GX$ is tangent to $(LDX)$ Hence $\angle GXB=\angle AKL=\angle BKX\Rightarrow GX$ is tangent to $(KEBX)$ Thus $(BEK)$ is tangent to $(DFL)$ at $X.$

Denote the reflection of $A$ across $BC$ by $A'$ and let $AA'$ intersect $BC$ and $CD$ again at $X$ and $Y$. The key claim is the following. Claim: $EKBXA'$ and $FLDYA'$ are cyclic. Proof. Note that $B$ is the orthocenter of $\triangle AKX$. A homothety with scale factor $2$ at $A$ sends the nine point circle of $AKX$ to $(EKBXA')$ and symmetry implies the result. $\blacksquare$ Now, if $\ell$ is the line through $A'$ tangent to $(EKXBA')$, we have \[\measuredangle (\ell, \overline{AXY})=\measuredangle A'BX=\measuredangle A'BC=\measuredangle A'DC=\measuredangle A'DY\]where the fourth equality follows from $A'BCD$ being cyclic fro the given angle condition. This finishes.

Let $X$ be the reflection of $A$ over $BC$, then $BXCD$ is cyclic. I claim that $X$ is the tangency point (motivated by reflections about two of the sides of $\triangle BCD$, it's worth looking at the last reflection too to get some circumcenters). We use directed angles mod $180^\circ$.

Let $T$ be the reflection point of $A$ across $BD$. Note that $BE$ and $BT$ are just reflections of $BA$ across $BC$ and $BD$ respectively, so $\angle BEK = \angle BAK = \angle BTK$ and thus $K, E, T, B$ are concyclic, so similarly $L, F, T, D$ are too. It suffices to prove the orientation of the line tangent to $T$ from $(KEB)$ (say $l_1$) is the same as those from $(LFD)$ (say $l_2$). Indeed, using formal sum for angle chasing we see $$ KB + l_1 = KT + TB, DL + l_2 = LT+TD. $$$KB=DL$ are the same line, so $l_1 = l_2$ iff $(KT-LT) + (BT-DT) = 0$. Indeed, this equates to $(2KB - KA + 2KB - AB) - (2KB - LA + 2KB - DA)$ $$ = LA + DA - KA - AB = (LA - KA) + (DA - AB) = \sphericalangle KAL + \sphericalangle BAD = \sphericalangle KAL - \sphericalangle BCD = 0,$$where the last equation follows by the fact that $A, B', C, D'$ is concyclic.

Let $T$ be the point on line $BD$ for which $\overline{TA}$ is tangent to $(ABK)$. Then, angle chasing gives us $\angle TAD = \angle DLA$, so $\overline{TA}$ is also tangent to $(ADL)$. Therefore, $(ABK)$ and $(ADL)$ are tangent at $A$. Since $\angle KEB = 180^{\circ} - \angle KAB$ and $\angle LFD = 180^{\circ} - \angle LAD$, it follows that $(KEB)$ and $(DFL)$ are the reflections of $(ABK)$ and $(ADL)$ over line $BD$, respectively, so we have the desired tangency.

We claim that the tangency point is the reflection of $A$ across $BD$. Call this point $T$. Claim: $T$ lies on $(KEB)$. Note that $BE=BA$, but $BT=BA$ as well, so $B$ is the circumcenter of $AET$ and $BE=BT$. However, we have $\angle EKB=\angle TKB$ by symmetry, so $B$ is the arc midpoint of $ET$. Similarly, $T$ lies on $(LFD)$ as well. Thus, it suffices to show that they are tangent. Let $\angle BET=\angle BTE=\angle BKE=\angle BKT=\theta_b$, and define $\theta_d$ similarly. It suffices to show that $\angle BTD=\angle BAD=\theta_b+\theta_c$, since then the line through $T$ tangent to $(KEBT)$ would also be tangent to $(LFDT)$ as well. This is simply a matter of angle chasing. We have $$180-\theta_b-\theta_c=\angle KAL=\angle KAB+\angle BAD+\angle LAD$$$$=90-(180-\angle ABT)+\angle BAD+90-(180-\angle ADT)$$$$=\angle BAD+\angle ABT+\angle ADT-180=\angle BAD+(360-2\angle BAD)-180=180-\angle BAD,$$as desired.

Reflect $A$ about $BD$ to a point $A'$, and note that $A'BDC$ is cyclic. Claim: $(BEK)$ passes through $A'$. Proof. Immediately follows from $$\angle BA'K = \angle BAK = \angle BEA$$proving the claim. $\square$ Likewise $(DFL)$ passes through $A'$. Now let $T$ on $(A'BDC)$ be chosen so that $CT \perp BD$. Claim: $A'T$ is the common internal tangent to $(BEK)$ and $(DFL)$. Proof. Proceed with complex numbers, setting $(A'BDC)$ as the unit circle. Then we compute, $t = -bd/c$ $a = b + d - bd/a'$ $e = b + c - bc\overline{a} = b - bc/d + a'c/d$ Then to verify $\angle TA'B = \angle A'EB$ we have to check, \begin{align*} \frac{t - a'}{b - a'} \div \frac{a' - e}{b - e} \in \mathbb{R} \end{align*}However this follows as, \begin{align*} \frac{t - a'}{b - a'} \div \frac{a' - e}{b - e} &= \frac{-bd/c - a'}{b - a'} \div \frac{a' - b + bc/d - a'c/d}{bc/d - a'c/d}\\ &= \frac{bd + a'c}{a'c - bc} \div \frac{a'd - bd + bc - a'c}{bc - a'c}\\ &= \frac{bd + a'c}{a'c - bc} \div \frac{(b - a')(c - d)}{c(b - a')}\\ &= \frac{bd + a'c}{(a' - b)(c - d)} \end{align*}which is clearly self conjugate. Thus both circles are tangent at $A'$ and we're done. $\square$

Notice that $(ABD)$ and $(CBD)$ are reflections of each other over $BD$, so that along with all the other reflections motivates us to reflect $A$ over $BD$, and we call this reflection $K$. By angle chasing, we show that $K$ lies on both the circles we are supposed to prove tangent, so it just remains to show that $K$ is the tangency point, which follows from some more angle chasing.

All one needs to spot is a single construction and the question falls apart. Reflect $A$ in $BD$ to get $A’$. Let $X,Y,Z$ be feet of perpendiculars on $DB,BC,CD$. CLAIM 1: $A’ \in \odot(BEK)$ PROOF: Note that $\angle A’EK=\angle A’EA=\angle XYA=\angle XBA=\angle A’BX$ and $\angle A’FL=\angle A’FA=\angle XZA=\angle XDA=\angle XDA’$ CLAIM 2: $\odot(BEK)$ is tangent to $\odot(DLF)$ PROOF: Consider a line $l$ passing through $A’$ tangent to $\odot(BEK)$ which meets $B$ in $U$. Observe that $\angle BA’U=\angle BKA’=\angle BKA$ and $\angle UA’D=\angle BA’D- \angle BA’U=\angle KLA+\angle LKA-\angle LKA=\angle KLA=\angle DLA’$, so $l$ is also tangent to $\odot(DFL)$ and we are done. $\blacksquare$

Consider $X$ the relection of $A$ across line $BD$. Claim 1: $X\in (BEK)\cap(DLF)$. Proof: Note that $\angle DFL=\angle DAL=\angle DXL$, hence $D,X, F,L$ are concyclic. Similarly, $K,X,B,E$ are concyclic, which leads to the desired answer. Claim 2: Point X is in fact the tangency point between the two circles. Proof: Let $S$ be the intersection of the tangent at $X$ to $(BEKX)$ with $BD$. All that remains is to prove that $XS$ is tangent to $(DLFX)$. We do so by angle chase. We have that $\angle BXS=\angle XKB$ implying that $\angle BAS=\angle BKA$. Denote $\angle BAD =\angle BCD = \alpha$. Then, $\angle DXS= \angle DAS=\alpha -\angle BAS = \alpha -\angle BKA$. We easily obtain that $\angle LAK = 180 - \alpha$, thus $\angle DLA=\angle DLX= 180 - (\angle BAK + 180 - \alpha)=\alpha -\angle BAK$. Therefore, $\angle DLX=\angle DXS$, leading to the conclusion.

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draw(circle((0.18461157234962958,13.13596824490276), 6.810781386792311), linewidth(1) + xfqqff); draw((xmin, 0.6364294452151896*xmin + 4.945351290292978)--(xmax, 0.6364294452151896*xmax + 4.945351290292978), linewidth(1) + xfqqff); /* line */ draw(circle((-2.007988314762082,3.322927456675997), 7.124455616338704), linewidth(1) + aqaqaq); draw((-5.570003624499658,9.493009689707842)--(3.8414223143281734,7.390145562638109), linewidth(1) + qqccqq); draw(arc((3.8414223143281734,7.390145562638109),1.7537677260641478,167.40486368764013,244.10032740676203), linewidth(1) + red); draw(arc((3.8414223143281734,7.390145562638109),1.4907025671545255,167.40486368764013,244.10032740676203), linewidth(1) + red); /* dots and labels */ dot((-12.271102218877283,2.558151842037977),dotstyle); label("$A$", (-12.077524302347232,2.9810044520348353), NE * labelscalefactor); dot((-5.570003624499658,9.493009689707842),dotstyle); label("$B$", (-5.41320694330347,9.952231163139817), NE * labelscalefactor); dot((5.105002275846094,3.7269481084991902),dotstyle); label("$C$", (5.284776185687833,4.164797667128134), NE * labelscalefactor); dot((-1.586833777894707,-3.789069216921096),linewidth(4pt) + dotstyle); label("$D$", (-1.4233853665075327,-3.4202477480992997), NE * labelscalefactor); dot((-3.4444736552568354,18.899342561173142),dotstyle); label("$E$", (-3.2648414788748883,19.334888497583), NE * labelscalefactor); dot((5.953220529990806,-13.667752748870427),dotstyle); label("$F$", (6.1178158555683035,-13.24134701405852), NE * labelscalefactor); dot((-6.625023716050909,13.011026908081407),linewidth(4pt) + dotstyle); label("$K$", (-6.465467578941958,13.372078228964904), NE * labelscalefactor); dot((-0.2917524693671866,-8.10758256001249),linewidth(4pt) + dotstyle); label("$L$", (-0.1080595719594218,-7.760822870108062), NE * labelscalefactor); dot((3.8414223143281734,7.390145562638109),linewidth(4pt) + dotstyle); label("$I$", (4.013294584291326,7.760021505559635), NE * labelscalefactor); dot((3.8414223143281743,7.39014556263811),dotstyle); label("$A'$", (4.013294584291326,7.847709891862842), NE * labelscalefactor); dot((-7.857787937067057,10.728747201605563),linewidth(4pt) + dotstyle); label("$Y$", (-7.693104987186862,11.092180185081514), NE * labelscalefactor); dot((-4.214839952274555,4.974148702338042),linewidth(4pt) + dotstyle); label("$X$", (-4.054036955603754,5.304746689069829), NE * labelscalefactor); dot((-3.1589408444432387,-5.554800453416224),linewidth(4pt) + dotstyle); label("$Z$", (-3.001776319965266,-5.21785966731505), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Let $A'$ be the reflection of $A$ over $BD$ and $X,Y,Z$ be the projections on $BD, BC, CD$, respectively. Lemma: $(A'BKE)$ and $(A'CLF)$. Proof: Both claims are analogous so we'll prove $(A'BKE)$. Notice that: $$\angle AEA' = \angle AYX = \angle ABX = \angle A'BX = 180^{\circ} - \angle A'BK \ _{_{\blacksquare}}$$Because $\angle BCD = \angle BAD = \angle BA'D$ we have $(BA'CD)$. Let $ \ \ell$ be the tangent to $(A'BKE)$ at $A'$. We have: $$\measuredangle (BA'; \ell) = \angle BKA' = \angle BKA = 90^{\circ} - \angle KAX = 90^{\circ} - \angle DBC.$$$$\measuredangle (DA'; \ell) = \angle BA'D - \measuredangle (BA'; \ell) = \angle BCD + \angle DBC -90^{\circ} = 90^{\circ} + \angle BDC = \angle DLA'.$$It implies that $\ell$ is tangent to $(A'CLF)$ and we are done.$ \ _{\blacksquare}$