Quite easy for a J4. Once you add the midpoints of $AB$ and $CD$ and realize there's a translation taking $ABK$ to $DCL$, you can use parallelograms and angle chase to finish. One Thing: I don't know if you can automatically assume the triangles are directly similar (based on the inside/outside condition on $K$ and $L$). This shouldn't be a big issue, however.

Let $AC \cap BD = (0,0)$, $A=(-a,0)$, $B=(0,b)$, $C=(a,0)$, and $D=(0,-b)$. Since $KA=KB$, $K$ lies on the perpendicular bisector of $AB$. The midpoint of $AB$ is $\left(\frac{-a}{2},\frac{b}{2}\right)$. $AB$ has slope $\frac{b}{a}$, therefore the perpendicular bisector of $AB$ has slope $\frac{-a}{b}$. By point-slope form, the equation of the perpendicular bisector of $AB$ is $y=\frac{-a}{b}\left(x+\frac{a}{2}\right)+\frac{b}{2}$. Similarly, we find that $L$ lies on the perpendicular bisector of $CD$ with equation $y=\frac{-a}{b}\left(x-\frac{a}{2}\right)-\frac{b}{2}$. Now, let $K=\left(x_1,\frac{-a}{b}\left(x_1+\frac{a}{2}\right)+\frac{b}{2}\right)$. Due to the length conditions, a vector transformation of $\overrightarrow{AD}$ maps $K$ to $L$. Therefore, $L=\left(x_1+a,\frac{-a}{b}\left(x_1+\frac{a}{2}\right)-\frac{b}{2}\right)$. The midpoint of $KL$ is $\left(x_1+\frac{a}{2},\frac{-a}{b}\left(x_1+\frac{a}{2}\right)\right)$. As $KL$ is parallel to $AD$, the perpendicular bisector of $KL$ has slope $\frac{a}{b}$. By point-slope form, the equation of the perpendicular bisector of $KL$ is $y=\frac{a}{b}\left(x-2x_1-a\right)$. It suffices to prove that the midpoint of the $x$ and $y$ intercepts is also the midpoint of $KL$, $\left(x_1+\frac{a}{2},\frac{-a}{b}\left(x_1+\frac{a}{2}\right)\right)$. Plugging in $0$ for $x$ and $y$ separately, the $x$-intercept is found to be $(a+2x_1,0)$, and the $y$-intercept is found to be $\left(0, \frac{-a\left(2x_1+a\right)}{b}\right)$. The midpoint of those points is $\left(x_1+\frac{a}{2},\frac{-a(2x_1+a)}{2b}\right)$, equivalent to the midpoint of $KL$, so we're done. $\blacksquare$

Sketch: Construct midpoints $R$ and $S$ of $AB$ and $CD.$ Let $O=AC \cap BD$ and $M$ the midpoint of $KL.$ You get two parallelograms in which then you can show $OM=XM=YM,$ which then it becomes trivial on how to finish. [asy][asy] import olympiad; unitsize(50); defaultpen(fontsize(8.5pt)); dotfactor*=1; pair A,B,C,D,O,K,L,X,Y,M,R,S; A = (-1,0); B = (0,2); C = (1,0); D = (0,-2); O = (0,0); K = (-0.2,0.85); L = (0.8,-1.15); X = (0.6,0); Y = (0,-0.3); M = midpoint(K--L); R = midpoint(A--B); S = midpoint(C--D); /* Drawing */ draw(A--B--C--D--cycle); draw(B--D); draw(A--C); draw(K--L); draw(Y--X); draw(R--S); draw(R--K); draw(S--L); draw(A--K); draw(B--K); draw(D--L); draw(C--L); draw(K--X--L--Y--cycle, dotted); draw(O--M); /* Labeling */ dot("$A$", A, W); dot("$B$", B, N); dot("$C$", C, E); dot("$D$", D, S); dot("$K$", K, N); dot("$L$", L, NE); dot("$X$", X, N); dot("$Y$", Y, W); dot("$M$", M, NW); dot("$R$", R, NW); dot("$S$", S, W); dot("$O$", O, SW); [/asy][/asy]

coordbash ftw lol

imagine not coordbashing. Let $M$ be the midpoint of $KL,$ let the perpendicular bisector of $KL$ meet $BD$ and $AC$ at $Y$ and $X$ respectively, I claim $KXLY$ is a rhombus. Lets proceed with coordinates. Set the center of $ABCD$ be $(0, 0),$ $A=(0, 2b),$ and $B=(-2a, 0).$ So, $C=(0, -2b),$ $D=(2a, 0).$ Let $M_1$ be the midpoint of $AB$ and $M_2$ the midpoint of $CD,$ we have $M_1=(-a, b), M_2=(a, -b).$ Now, look at line $KL,$ since $\triangle{ABK} \cong \triangle{DCL},$ $M_1M_2LK$ is a parallelogram and the slope of line $M_1M_2$ is $-\frac{b}{a},$ so, the slope of line $KL$ is $-\frac{b}{a}$ by parallel lines. Since $M_2L$ is perpendicular to line $CD,$ the slope of line $M_2L$ is $-\frac{a}{b}.$ So, the equation of line $M_2L$ is $y=-\frac{a}{b}x+n,$ for some $n,$ plugging in $(a, -b)$ gives $$-b=-\frac{a^2}{b}+n \implies n=\frac{a^2-b^2}{b}.$$Now plug in any $x$ call it $c$ for point $L$ to get $L=(c, \frac{a^2-b^2-ac}{b}).$ We can do similar things on triangle $ABK$ to get $K=(-2a+c, \frac{a^2+b^2-ac}{b}),$ we know the $x$ coordinate because from $M_2$ to $L$ we added $c-a$ to the $x$ coordinate, so $-a+c-a=-2a+c.$ From $K$ and $L,$ we get that the midpoint $M$ of $KL$ is $(-a+c, \frac{a^2-ac}{b}).$ Now, notice that line $XY$ has a slope of $\frac{a}{b},$ so the equation of $XY$ is $y=\frac{a}{b}x+n$ for some $n,$ plugging in $(-a+c, \frac{a^2-ac}{b})$ gives $$\frac {-a^2+ac}{b}+n=\frac{a^2-ac}{b} \implies n=\frac{2a^2-2ac}{b}.$$So, the line $XY$ has equation $y=\frac{a}{b}x+\frac{2a^2-2ac}{b}.$ Now, the equation of line $BD$ is $y=0,$ so, the $x$ coordinate for the point of intersection of $BD$ and $XY$ is $$\frac{a}{b}x+\frac{2a^2-2ac}{b}=0 \implies x=2c-2a.$$So, $y=(2c-2a, 0).$ Do similar things to get $X=(0, \frac{2a^2-2ac}{b}).$ So, $X=(0, \frac{2a^2-2ac}{b}), y=\frac{a}{b}x+\frac{2a^2-2ac}{b},$ and $M=(-a+c, \frac{a^2-ac}{b}).$ But the midpoint of $X$ and $Y$ is $(-a+c, \frac{a^2-ac}{b}),$ so $M$ is the midpoint of $XY.$ From here, its trivial to see that $KXLY$ is a rhombus.

First, I claim $\overline{KL} \parallel \overline{AD} \parallel \overline{BC}$. This is true as $\triangle KAB \sim \triangle LCD$ obviously, so $\angle LDA + \angle DAK = \angle CDA + \angle BAD = 180^\circ$, implying $\overline{KL} \parallel \overline{AD}$. So $\overline{KL}\parallel \overline{BC}$ as well. Next, define $X$ and $Y$ to be the intersections of the circle at $K$ through $A, B$ and the circle at $L$ through $C, D$. These intersection points exist because $$r_K+r_L = AK+BK > AB = AD = KL$$by the triangle inequality. Evidently, the circles are congruent, so $KYLX$ is a rhombus. It remains to show that $Y \in \overline{BD}$ and $X \in \overline{AC}$. Claim. $\triangle AKB \cong\triangle KYL \cong \triangle KXL \cong \triangle DLC$. Proof. First $\overline{KL} \parallel \overline{AD} \parallel \overline{BC}$ by the first claim. Then, as $YK=YL=AK=BK$ by construction and $KL=AD=AB$, $\triangle KYL \cong \triangle KAB$ by SSS. The other congruences follow similarly. $\blacksquare$ Thus $\angle YKL = \angle KAB$, and it follows $\angle AKB + \angle YKX = 180^\circ$, implying $\angle AKY + \angle BKX = 180^\circ$. As $\angle BKL = 180^\circ - \angle KLC$, $$\angle AKY = \angle KLC + \angle LKX = \angle KLC + \angle YLK = \angle YLC.$$So by SAS, $\triangle YLC \cong \triangle AKY$, so $AY=CY$. Thus, $Y$ lies on the perpendicular bisector of $\overline{AC}$, or $\overline{BD}$. Similarly $X$ lies on $\overline{AC}$ as well, so we are done.

Let $KA = x$ for convenience. Let $X, Y$ be the other points on $AC, BD$ such that $x=LX=LY$ (i.e. the second intersections of the circle centered at $L$ with radius $x$). Observe that since $\measuredangle (AD, DY) = \measuredangle BDC$ and $\measuredangle (DY, XY) = \measuredangle DYX = \measuredangle DCX,$ it follows that $XY\perp AD\parallel BC.$ Also, $\measuredangle BYX = \measuredangle DCA = \measuredangle BAX$ so $ABXY$ is cyclic. Finally, since $\triangle AKB\cong\triangle DLC,$ we know that $AD\parallel KL\parallel BC,$ whence $K$ also lies on the perpendicular bisector of $XY.$ This implies $K$ is the center of $(ABXY)$ and $KXLY$ is a rhombus with side length $x$ as desired. $\blacksquare$

Here's a diagram that I spent way too long drawing:

trivial by noticing that K and L can only vary along parallel, equal length line segments

its an easy coordinate bash

can't you just wlog b=2? (since dilations don't affect anything)

channing421 wrote:

can't you just wlog b=2? (since dilations don't affect anything) Sure, that does make it easier. On that note, this isn't a formal writeup or anything, just how I solved it on paper, which is why it's not too well written (ex: I used x_1 when there wasn't even an x_2)

almost forgot to show that the perpendicular bisector of KL actually intersects both diagonals (ie they are not parallel) noticed in like the last 30 minutes

I've got a coordbash that is slightly less computationally intensive, I think.

Did anyone else use hybrid synthetic + coords? Let $O$ be the intersection of diagonals $AC$ and $BD$, let $M$, $N$, and $P$ be the midpoints of $AB$, $CD$, and $KL$, respectively, and let $d = \frac{OB}{OA}$. Finally, let the perpendicular bisector of $KL$ intersect lines $AC$ and $BD$ at $X$ and $Y$, respectively. We claim $KXLY$ is a rhombus. Stick everything on the coordinate plane so that $AC$ and $BD$ coincide with the $y-$ and $x-$ axes, respectively. Note that the translation $T$ that maps $M$ to $K$ also maps $N$ to $L$, so $T$ maps $O$ to $P$ and $OP\perp AB$. In particular, we can write $P = (x_0,dx_0)$. Then the perpendicular bisector of $KL$ has equation $y = -dx + 2dx_0$. But $X$ and $Y$ are the $x-$ and $y-$ intercepts of line $KL$, respectively, so $X = (2x_0,0)$ and $Y = (0,2dx_0)$. It follows that $P$ is the midpoint of $XY$, so the diagonals of $KXLY$ are perpendicular bisectors of each other, as desired.

as somebody with a 174 jmo index who has solved 2 jmo problems before (2011 and 2018 p1 lmao), i'd like to thank maa for letting me say "i can do olympiad geo"

im dum. I got stuck in the middle and just proved some hard thing by saying "by symmetry, this must be true". Therefore, I probably got a 2 or something.

rip file size too big

CT17 wrote: asdf334 wrote: Zorger74 wrote: I'm transcribing as written from the solution I submitted during the test. Does it get a 7? I dunno asy, so I'm leaving out the first-page-diagram.

at mos 6 since you have to show perp bisector of KL actually intersects AC,BD rip Are you sure you need to prove this? Isn't it almost trivial that if the perpendicular bisector of $KL$ was parallel to $AC$ or $BD$ then the rhombus would be degenerate? By the same reasoning one could argue that you need to prove triangles $KAB$ and $LDC$ are directly similar (as mentioned in post 1), or that you need to prove $\frac{g_2}{g_1}$ is an integer on problem 1. I think you don't need to prove it if you coordbash, as if you set $(a,0), (-a, 0), (0, b), (0, -b)$ as the points of the rhombus, then the fact that $a, b\ne 0$, and then the fact that you calculate to get specific values for coordinates of intersection points would mean that it would have to intersect... correct me if im wrong

CT17 wrote: asdf334 wrote: Zorger74 wrote: I'm transcribing as written from the solution I submitted during the test. Does it get a 7? I dunno asy, so I'm leaving out the first-page-diagram.

at mos 6 since you have to show perp bisector of KL actually intersects AC,BD rip Are you sure you need to prove this? Isn't it almost trivial that if the perpendicular bisector of $KL$ was parallel to $AC$ or $BD$ then the rhombus would be degenerate? By the same reasoning one could argue that you need to prove triangles $KAB$ and $LDC$ are directly similar (as mentioned in post 1), or that you need to prove $\frac{g_2}{g_1}$ is an integer on problem 1.

CT17 wrote: asdf334 wrote: Zorger74 wrote: I'm transcribing as written from the solution I submitted during the test. Does it get a 7? I dunno asy, so I'm leaving out the first-page-diagram.

at mos 6 since you have to show perp bisector of KL actually intersects AC,BD rip Are you sure you need to prove this? Isn't it almost trivial that if the perpendicular bisector of $KL$ was parallel to $AC$ or $BD$ then the rhombus would be degenerate? By the same reasoning one could argue that you need to prove triangles $KAB$ and $LDC$ are directly similar (as mentioned in post 1), or that you need to prove $\frac{g_2}{g_1}$ is an integer on problem 1. I did prove that g2/g1 was an integer LOL

Let $A=(0,-2b)$, $B=(2a,0)$, $C=(0,2b)$, $D=(-2a,0)$. Since $KA=KB=LC=LD$, $K,L$ lie on the perpendicular bisectors of $AB,CD$ respectively. So we can write: $$K=\left(k,-\frac{ak}b+\frac{a^2}b-b\right),L=\left(\ell,-\frac{a\ell}b-\frac{a^2}b+b\right)$$for some constant $k,\ell$. Let $k+\ell=j$, then the midpoint of $KL$ is: $$J=\left(\frac j2,\frac{-aj}{2b}\right).$$ On the other hand, note that $LKBC$ is a parallelogram, so the slope of $KL$ is the slope of $CB$, $-\frac ba$. Then the perpendicular bisector of $KL$ is $y=\frac abx-\frac{aj}b$. The $x$- and $y$- intercepts are $\left(0,-\frac{aj}b\right)$ and $(j,0)$, so since their midpoint is $J$, so $KXLY$ is indeed a rhombus.

Realized I never posted this. It's a bit janky, but I don't see it in this thread so whatever. Let $M$ be the midpoint of $\overline{KL}$. Choose points $X\in AC$ and $Y\in BD$ such that $XY$ is the perpendicular bisector of $\overline{KL}$. We claim $MX=MY$; this will imply $KXLY$ is a rhombus. [asy][asy] size(250); defaultpen(linewidth(0.6)+fontsize(10)); real r = 0.13; pair A = (0,0), B = (3,4), C = (8,4), D = (5,0), M1 = (A+B)/2, M2 = (C+D)/2; pair K = M1 + (4*r,-3*r), L = M2 + (4*r,-3*r), P = (M1+M2)/2, M=(K+L)/2; pair Xp = rotate(90,M)*L, X = extension(M,Xp,A,C), Y = extension(M,Xp,B,D), T = extension(M,Xp,B,C); draw(A--B--C--D--A--C^^B--D,rgb(0.8,0.1,0.1)); draw(A--K--B^^C--L--D,rgb(0.9,0.5,0.1)); draw(M1--K--L--M2,rgb(0.1,0.6,0.1)); draw(X--Y,rgb(0.1,0.1,0.7)); draw(X--T,rgb(0.1,0.1,0.7)+linetype("3 3")); draw(circumcircle(D,C,X),gray(0.6)+linetype("4 4")); clip((-0.5,-0.5)--(-0.5,4.5)--(8.5,4.5)--(8.5,-0.5)--cycle); dot("$A$",A,SW,linewidth(3.3)); dot("$B$",B,NW,linewidth(3.3)); dot("$C$",C,NE,linewidth(3.3)); dot("$D$",D,SE,linewidth(3.3)); dot("$K$",K,SE,linewidth(3.3)); dot("$L$",L,SE,linewidth(3.3)); dot("$M_1$",M1,NW,linewidth(3.3)); dot("$M_2$",M2,NW,linewidth(3.3)); dot("$M$",M,SE,linewidth(3.3)); dot("$X$",X,NW,linewidth(3.3)); dot("$Y$",Y,SW,linewidth(3.3)); dot("$T$",T,N,linewidth(3.3)); dot("$P$",P,W); [/asy][/asy] Let $M_1$ be the midpoint of $\overline{AB}$, $M_2$ the midpoint of $\overline{CD}$, and $P$ be the intersection of the diagonals $AC$ and $BD$. Note that $P$ is the midpoint of $\overline{M_1M_2}$. Combined with $KM_1\parallel LM_2$, we thus obtain $MP\parallel KM_1$. That is, $MP$ is perpendicular to $AB$ and $CD$. Now write $T := XMY\cap BC$. Since $\angle YPC = \angle YTC = 90^\circ$, we have \[ \measuredangle DYX\equiv \measuredangle PYT = \measuredangle PCT = \measuredangle DCX, \]so quadrilateral $DXYC$ is cyclic. Thus, since $PM$ is an altitude of $\triangle CDP$, it passes through the circumcenter of $\triangle PXY$. That is, $M$ is the midpoint of $\overline{XY}$.

Let $A = (0, a)$ and $B = (b, 0)$ for positive $a$ and $b$, and let $K = (d, c)$. Since $KA = KB$, $K$ lies on the perpendicular bisector of $AB$. This means that $c - \frac{a}{2} = \frac{b}{a}(d - \frac{b}{2})$ Claim. For any fixed $K$, $L$ is unique Proof. Since $LC = LD$, $L$ lies on the perpendicular bisector of $CD$. Only two points on the perpendicular bisector satisfy $LC = KA$, which are $(d-b, c-a)$ and $(-d, -c)$. It isn't too hard to show that both work and are the only two. However, since $K$ is in the rhombus, so is $(-d, -c)$. Thus, the only $L$ which satisfies all conditions is $(d-b,c-a)$. Now I claim that the points $X = (2d-b, 0)$, and $Y = (0, 2c-a)$ satisfy the condition. Notice that the slope of line $XY$ is $\frac{a-2c}{2d-b} = -\frac{b}{a}$, and the slope of line $KL$ is $\frac{a}{b}$, so $KL$ and $XY$ are perpendicular. The midpoint of both segments is also $(d-\frac{b}{2},c-\frac{a}{2})$. Thus, diagonals $KL$ and $XY$ are perpendicular and bisect each other, meaning that $KXLY$ is a rhombus as desired.

Claim: $AD \parallel KL \parallel BC$ Proof: Notice that $\triangle KAB \cong \triangle LDC$, where $\angle LDC = \angle KAB$, then $\angle LDA + \angle DAK = \angle LDC + \angle CDA + \angle DAB - \angle KAB = 180^{\circ}$ (Since $ABCD$ Is a rhombus). Analogously $\angle LCB + \angle CBK = 180^{\circ}$ $\square$ Let $P$ the intersection of $AC$ with $BD$ and let $X$, $Y$ be intersections of the perpendicular bisector of $KL$ with $AC$ and $BD$, respectavely. Now, $\angle PYX = 90^{\circ} - \angle ACB = \angle BDC = \angle XDC \Rightarrow YXCD$ is cyclic y as $LD = LC$, thus $L$ is the center of that circle. Therefore, $KX = LX = LY = YK$, that is, $KXLY$ is a rhombus $\blacksquare$

Synthetic ftw. Let $X,Y$ be the intersection of $AC$ and $BD$ respectively, and the perpendicular bisector of $KL$. Let $M$ be the midpoint of $KL$. We wish to prove that $MX=MY$, which implies that $KXLY$ is a rhombus. Let $O=AC\cap BD$. We wish to prove that $MX=MO$, because from which $MY=MO$ can be implied similarly, which finishes the problem. Let $N,P$ be the midpoint of $AB$ and $CD$, respectively. Then $NP || AD$ and $NK || PL$, which shows that $NKLP$ is a parallelogram. It's obvious that triangles $OBN$ and $ODP$ are congruent, which implies that $N,O,P$ are on a line and $ON=OP$, which implies that $OM || NK$ and thus $OM \perp AB$. Therefore $\angle MOX=90^\circ - \angle BAC$. Now note that $MX \perp KL || BC$. Thus $\angle MXO=90^\circ -\angle BCA=90^\circ -\angle BAC=\angle MOX$. Thus triangle $MOX$ is isoseles at $M$ and we are done.

Let we denote by $S$ and $R$ the intersections of $(ABK)$ with lines $ AC $ and $BD$ respectively. Now we have $ABSR $ cyclic with circumcenter $K$. $ \angle DCS = \angle CAB = \angle SAB = \angle SRB \ \Rightarrow$ $RDSC$ is cyclic. Let $O$ be the circumcenter of $RDSC$. We will wont to show that $L \equiv O. $ We know $LD = LC$. So $ L$ and $O$ are on the perpendicular bisector of $DC$. $\angle ODC = \angle OCD = \frac {180 - \angle DOC}{2} = \frac {180 - 360 + 2\angle DSC}{2} = \angle DSC - 90 $ $\angle KAB = \frac {180 - \angle AKB}{2} = 90 - \angle ASB $ But $AC$ is perpendicular bisector of $BD$ $\Rightarrow$ $ \angle ASD = \angle ASB $ $\Rightarrow$ $\angle ASB + \angle DSC = \angle ASD + \angle DSC = 180$ $\Rightarrow$ $ \angle ODC = \angle DSC - 90 = 180 - \angle ASD - 90 = 90 - \angle ASD = \angle KAB $ So we have $ O \equiv L $. Now $AK = KB = LD = LC = LS = LR$ $\Rightarrow$ $KSLR$ is a rhombus.

We employ cartesian coordinates, with $A = (0,1), B = (a,0), C = (0,-1),$ and $D = (-a,0)$ for some real number $a > 0.$ Note that $\triangle AKB \cong \triangle CLB$ and are oriented the same direction, hence $L-K=D-A$. Now we get to the computation. Note that $K$ is an arbitrary point on the perpendicular bisector of segment $AB.$ The midpoint of $AB$ is $\left(\frac{a}{2}, \frac{1}{2}\right)$ and the slope of line $AB$ is $-\frac{1}{a}.$ Thus the slope of the perpendicular bisector of $AB$ is $-\frac{1}{-\frac{1}{a}} = a.$ The slope and fixed point then tells us that the perpendicular bisector $\ell$ of $AB$ has equation $y = ax + \frac{1}{2} - \frac{a^2}{2}.$ Thus let $K = \left(x_1, ax_1 + \frac{1}{2} - \frac{a^2}{2}\right)$ for real $x_1 \ne \frac{a}{2}$ (otherwise $K$ is the midpoint of $AB$) so that $L = K + D - A = \left(x_1-a, ax_1 - \frac{1}{2} - \frac{a^2}{2}\right).$ Now, let $X = (0,2ax_1 - a^2)$ and $Y = (2x_1 - a, 0).$ (Both of these are not the origin since $2x_1 - a \ne 0.$) Note that $X$ lies on the $y$-axis, which is line $AC,$ and $Y$ lies on the $x$-axis, which is line $BD.$ We claim that $KXLY$ is a rhombus, which will solve the problem. First, we must check that $KXLY$ is a parallelogram. This is easy; both segment $KL$ and $XY$ can easily be shown (e.g. by applying the midpoint formula to both segments) to share a midpoint \[ \left(x_1 - \frac{a}{2}, ax_1 - \frac{a^2}{2}\right). \]All that is left to show is that $XY \perp KL.$ Since $L-K = D-A,$ the slope of line $KL$ is the same as the slope of line $AD,$ which is $\frac{0-1}{-a-0} = \frac{1}{a}.$ Therefore, we want to show that the slope of $XY$ is $-a.$ This is easy enough; by the slope formula, we get \[ m_{XY} = \frac{0-(2ax_1 - a^2)}{(2x_1-a) - 0} = -\frac{2ax_1 - a^2}{2x_1 - a} = -\frac{a(2x_1-a)}{2x_1-a} = -a. \]Therefore, $XY \perp KL$ and share a common midpoint, which is enough to imply that $KXLY$ is a rhombus, as desired. Remark: I found $X$ and $Y$ by directly finding the perpendicular bisector of $KL$ and letting it hit the $x$ and $y$ axes.

:skull: Let the perpendicular bisector of $KL$ meet $AC$ and $BD$ at $X$ and $Y$, respectively, and let $KL$ itself meet $AC$ at $Z$. We reduce as follows: $KXLY$ is a rhombus. $M_{KL}X=M_{KL}Y$. $\angle M_{KL}ZX=\angle M_{KL}ZY$. Back to the definitions: $KM_{KL}L$ is a translation of $M_{AB}M_{AC}M_{CD}$ in the direction perpendicular to $AB$. Hence: $KL\parallel BC$ and so $\angle M_{KL}ZX=\angle ACB$. $M_{AC}M_{KL}\perp CD$ and so $\angle M_{KL}ZY=\angle M_{KL}M_{AC}Y$ (as $M_{AC},M_{KL}$ lie on the circle of diameter $YZ$) $=90^\circ-\angle CM_{AC}M_{KL}=\angle ACD$. Since $\angle ACB=\angle ACD$, we are done. $\square$

We use barycentric coordinates with $A=(1,0,0)$, $B=(0,1,0)$, and $C=(0,0,1)$. Note that $a=c$ and $D=A+C-B=(1,-1,1)$. Let $K=:(k_1,k_2,k_3)$. Since $\frac{1}{2}(A+B)=\left(\frac{1}{2},\frac{1}{2},0\right)$, we have \[ \left(k_1-\frac{1}{2},k_2-\frac{1}{2},k_3\right)\perp(1,-1,0) \]so \[ -a^2k_3+b^2k_3+a^2(k_2-k_1)=0\implies k_3=\frac{a^2}{b^2}(k_1-k_2+k_3). \]We also have \[ L=K+\overrightarrow{BC}=(k_1,k_2-1,k_3+1) \]so \[ M:=\frac{1}{2}(K+L)=\left(k_1,k_2-\frac{1}{2},k_3+\frac{1}{2}\right). \]Let $X=:(x_1,0,x_2)$ be the intersection of the perpendicular bisector of $\overline{LK}$ with $\overleftrightarrow{AC}$. Since $\overrightarrow{MX}\perp\overrightarrow{LK}$, we have \[ \left(x_1-k_1,-k_2+\frac{1}{2},x_2-k_3-\frac{1}{2}\right)\perp(0,1,-1) \]so \begin{align*} 0&=a^2\left(x_2-k_3-\frac{1}{2}+k_2-\frac{1}{2}\right)+b^2(k_1-x_1)+a^2(x_1-k_1)\\ &=a^2(x_1+x_2-1)-b^2k_3+b^2(k_1-x_1)\\ &=b^2(k_1-k_3-x_1)\\ \implies x_1=k_1-k_3. \end{align*}Let $Y=:(y_1,y_2,y_1)$ be the intersection of the perpendicular bisector of $\overline{LK}$ with $\overleftrightarrow{AC}$. Since $\overrightarrow{MY}\perp\overrightarrow{LK}$, we have \[ \left(y_1-k_1,y_2-k_2+\frac{1}{2},y_1-k_3-\frac{1}{2}\right)\perp(0,1,-1) \]so \begin{align*} 0&=a^2\left(y_1-k_3-\frac{1}{2}-y_2+k_2-\frac{1}{2}\right)+b^2(k_1-y_1)+a^2(y_1-k_1)\\ &=a^2(2y_1-y_2-k_1+k_2-k_3-1)+b^2(k_1-y_1)\\ &=a^2(4y_1-2)-b^2k_3+b^2(k_1-y_1)\\ \implies y_1=\frac{2a^2-b^2k_1+b^2k_3}{4a^2-b^2}. \end{align*}We also have \[ y_2=\frac{-b^2+2b^2k_1-2b^2k_3}{4a^2-b^2}. \]It suffices to check that $\frac{1}{2}(X+Y)=M$, or $x_1+y_1=2k_1$ and $y_2=2k_2-1$. We have \begin{align*} x_1+y_1&=\frac{2a^2-2b^2k_1+2b^2k_3+4a^2k_1-4a^2k_3}{4a^2-b^2}\\ &=2k_1+\frac{2a^2+2b^2k_3-4a^2k_3-4a^2k_1}{4a^2-b^2}\\ &=2k_1+\frac{2b^2k_3-2a^2k_3-2a^2k_1+2a^2k_2}{4a^2-b^2}\\ &=2k_1 \end{align*}and \begin{align*} y_2&=2k_2-1+\frac{-2b^2+4a^2+2b^2k_1+2b^2k_2-2b^2k_3-8a^2k_2}{4a^2-b^2}\\ &=2k_2-1+\frac{4a^2-4b^2k_3-8a^2k_2}{4a^2-b^2}\\ &=2k_2-1+\frac{-4b^2k_3+4a^2k_1-4a^2k_2+4a^2k_1}{4a^2-b^2}\\ &=2k_2-1 \end{align*}so we are done. $\square$

Simple and synthetic! Take $X,Y$ on $AC,BD$ such that $KA=KB=KY=KX$, Then $AYXB$ is cyclic since K is the center of $AYXB$. Then we have: $\angle DXA=\angle BXA=\angle AYB=\angle AYD$ which since $DB\perp AX$ implies that $DX\perp AY$ and then $D$ is the orthocenter of $\triangle AYX$ Since $K$ is the circumcenter of $AYX$ by definition and $ADLK$ is a parallelogram, we have that $KM=ML$.(Since By lemma $AD$=$2KM$) Thus $XY$ is the bisector of $KL$ and $XK=XL=LY=KY$

im bad. Let's view this problem in a more constructive light. Consider $ABCD$ rhombus with $BC$ as the $x$-axis. Consider $K$ and $L$ on the perpendicular bisectors of $AB$ and $CD$ respectively such that $KL$ is parallel to $BC$. Let the perpendicular bisector of $KL$ intersect $BD$ and $AC$ at $E$ and $G$, and the midpoint of $KL$ be $X$. We need to prove that $EX = XG$. Convince yourself this is the same as the problem. Proceed using coordinates, let $B=(0, 0), A=(a, b), C=(s, 0), D=(s+a, b)$ where $s^2=a^2+b^2$. The perpendicular bisector of $AB$ is $$y - \frac{b}{2} = -\frac{a}{b} \left( x-\frac{a}{2} \right)$$so consider fixing the $y$ coordinate of $KL$ as $y_0$. Then $$K = (-\frac{b}{a} (y_0 - \frac{b}{2}) + \frac{a}{2}, y_0)$$Shifting this to the right by $s$ gives us $$K = (-\frac{b}{a} (y_0 - \frac{b}{2}) + \frac{a}{2}+s, y_0)$$We can now find $X$, the midpoint of $KL$ as $$X = (-\frac{b}{a}(y_0 - \frac{b}{2})+\frac{a}{2}+\frac{s}{2}, y_0)$$So the perpendicular bisector of $KL$ has equation $x = X_x$. Intersecting this with $BD \Rightarrow y=\frac{b}{s+a} x$ and $AC \Rightarrow y = \frac{b}{a-s}(x-s)$ gives us $$E = (-\frac{b}{a}(y_0 - \frac{b}{2})+\frac{a}{2}+\frac{s}{2}, \frac{b}{s+a}(-\frac{b}{a}(y_0 - \frac{b}{2})+\frac{a}{2}+\frac{s}{2}))$$$$G = (-\frac{b}{a}(y_0 - \frac{b}{2})+\frac{a}{2}+\frac{s}{2}, \frac{b}{a-s}(-\frac{b}{a}(y_0 - \frac{b}{2})+\frac{a}{2}-\frac{s}{2}))$$We wish to prove that the midpoint of $EG$ is $X$, so it suffices to show that $$\frac{\frac{b}{s+a}(-\frac{b}{a}(y_0 - \frac{b}{2})+\frac{a}{2}+\frac{s}{2}) + \frac{b}{a-s}(-\frac{b}{a}(y_0 - \frac{b}{2})+\frac{a}{2}-\frac{s}{2})}{2} = y_0$$$$\frac{b(a^2+b^2-2by-s^2)}{(a-s)(a+s)} = 2y$$$$\frac{-2by^2}{a^2-s^2} = 2y$$$$-b^2=a^2-s^2$$, done.