Hmm nice! Let $ME \cap CK$ at $X$. Note that $\triangle CXM \equiv \triangle BXM \implies CX=XB$ and combined with $\angle XCB=60^\circ$ gives that $CXB$ is equilateral. Now, note that $\angle XKT = \angle KCB + \angle KBC = 75^\circ = \angle TEX$, so $KX=EX$ and thus $T,D,X$ are collinear. Finally, $\angle TXE=\angle TBD= 15^\circ$ so $XDEB$ is concyclic. Thus, $\angle TDE = \angle TBX = 45^\circ$. Now, $\angle KDE = 2 \times \angle TDE = 90^\circ$

This one was proposed by me

gghx wrote: Hmm nice! Let $ME \cap CK$ at $X$. Note that $\triangle CXM \equiv \triangle BXM \implies CX=XB$ and combined with $\angle XCB=60^\circ$ gives that $CXB$ is equilateral. Now, note that $\angle XKT = \angle KCB + \angle KBC = 75^\circ = \angle TEX$, so $KX=EX$ and thus $T,D,X$ are collinear. Finally, $\angle TXE=\angle TBD= 15^\circ$ so $XDEB$ is concyclic. Thus, $\angle TDE = \angle TBX = 45^\circ$. Now, $\angle KDE = 2 \times \angle TDE = 90^\circ$ @above is this the intended solution?

gghx wrote: Hmm nice! Let $ME \cap CK$ at $X$. Note that $\triangle CXM \equiv \triangle BXM \implies CX=XB$ and combined with $\angle XCB=60^\circ$ gives that $CXB$ is equilateral. Now, note that $\angle XKT = \angle KCB + \angle KBC = 75^\circ = \angle TEX$, so $KX=EX$ and thus $T,D,X$ are collinear. Finally, $\angle TXE=\angle TBD= 15^\circ$ so $XDEB$ is concyclic. Thus, $\angle TDE = \angle TBX = 45^\circ$. Now, $\angle KDE = 2 \times \angle TDE = 90^\circ$ When was T defined

Let $N$ be the point of intersection of $AC$ and the perpendicular bisector of $EK$ $\angle{KEN}=\angle{EKN}=75^{\circ}$ $\angle{KEM}=30^{\circ}+75^{\circ}=105^{\circ}$ Therefore $M,E,N$ are collinear $\angle{NAB}=\angle{NMB}=90^{\circ}$ so $AMBN$ is cyclic and since $CA=CM$ , then it,s a trapezoid , and so is $DEBN$ because $\angle{EBD}=\angle{DNE}$ and $\Delta ADN\cong \Delta BEM$ so $\angle{KED}=\angle{EBN}=45^{\circ}$ and we are done.

WLOG, let $AC=1$, then $BK=\frac{1}{\sin 15^{\circ}}=2\sqrt{3}\sqrt{2-\sqrt{3}}$, $BE=\frac{1}{\cos 15^{\circ}}=2\sqrt{2-\sqrt{3}}$. Let $X$ be the point of intersection between the perpendicular bisector of $EK$ and $EK$, if we can prove that $EX=DX=KX$, then it is clear that $X$ is the center of the circumcenter of triangle $EDK$ and $\angle EDK=90^{\circ}$. We have $EX=KX=\frac{BK-BE}{2}=(\sqrt 3 -1)\sqrt{2-\sqrt{3}}$ and also, $DX=\tan 15^{\circ}\times BX=(2-\sqrt 3)(\sqrt 3 +1)\sqrt{2-\sqrt 3}=(\sqrt 3 -1)\sqrt{2-\sqrt{3}}=EX=KX$. $\blacksquare$

$\begin{array}{l} DK = DE\mathop {}\limits_{}^{} ,\mathop {}\nolimits_{}^{} NC = NB\mathop {}\limits_{}^{} \Rightarrow \mathop {}\nolimits_{}^{} tr.NBC\mathop {}\limits_{}^{} is\mathop {}\limits_{}^{} equilateral\mathop {}\nolimits_{}^{} \Rightarrow \\ N\hat PB = 120^O \mathop {}\limits_{}^{} ,\mathop {}\limits_{}^{} and\mathop {}\nolimits_{}^{} tr.NPD = tr.EPB\mathop {}\nolimits_{}^{} \Rightarrow \mathop {}\nolimits_{}^{} \\ PD = PE\mathop {}\limits_{}^{} \Rightarrow \mathop {}\nolimits_{}^{} P\hat DE = 30^O \mathop {}\limits_{}^{} \Rightarrow \mathop {}\nolimits_{}^{} E\hat DH = 45^O \mathop {}\limits_{}^{} \Rightarrow \mathop {}\nolimits_{}^{} E\hat DK = 90^0 \\ \end{array}$

Let $X$ be on $AB$ so that $EX\perp AB$. Let $Y$ be the midpoint of $EK$. Since $BE$ is an angle bisector, $BX=BM$, and so since $EX\parallel AK$, $BE:EK=BX:BA=1:\sqrt{3}$ This implies $BY:BK=2+\sqrt{3}:1$, which is equal to $\tan{75}=\tan{\angle BKA}$, implying $AY$ is the angle bisector of angle $A$. Since $ADYK$ both cyclic (can be confirmed with right angles), we know $\angle YKD=\angle YAD=45$, implying $EDK$ is a right isosceles triangle. [asy][asy] import olympiad; pair A,B,C,M,E,K,X,Y,D; A=(0,-sqrt(3)); B=(0,0); C=(1,-sqrt(3)); M=C/2; K=extension(B,dir(-75),A,C); X=reflect(B,K)*M; E=sqrt(3)/3*K; Y=midpoint(E--K); D=rotate(90)*(E-Y)+Y; draw(A--B--C--cycle); draw(B--K); draw(X--E--M); draw(E--D--K); draw(D--Y); label("$A$",A,SW); label("$B$",B,N); label("$C$",C,SE); label("$M$",M,NE); label("$E$",E,dir(75)); label("$D$",D,W); label("$X$",X,W); label("$Y$",Y,dir(0)); label("$K$",K,S); [/asy][/asy]

Mikeglicker wrote: gghx wrote: Hmm nice! Let $ME \cap CK$ at $X$. Note that $\triangle CXM \equiv \triangle BXM \implies CX=XB$ and combined with $\angle XCB=60^\circ$ gives that $CXB$ is equilateral. Now, note that $\angle XKT = \angle KCB + \angle KBC = 75^\circ = \angle TEX$, so $KX=EX$ and thus $T,D,X$ are collinear. Finally, $\angle TXE=\angle TBD= 15^\circ$ so $XDEB$ is concyclic. Thus, $\angle TDE = \angle TBX = 45^\circ$. Now, $\angle KDE = 2 \times \angle TDE = 90^\circ$ When was T defined $T$ is the midpoint of $EK$

Note: It is assumed that the reader has chased all angles before this solution starts.

I found the mistake. The error was that I'd written $\sin 75 = \frac{BE}{4}$ instead of the correct $\sin 75 = \frac{4}{BE}$. The solution has been corrected.

Let ME meet CA at S. CB/2 = CM = CS/2 and ∠SCB = 60 so SCB is regular triangle. ∠ESK = 30 and ∠KES = 75 so SK = SE so ∠DSK = 15 = ∠DBK so DBSK is cyclic and ∠DKB = ∠DSB = 45 so ∠KDE = 90. we're Done.