If the calculations are ok then the answer is $ 60^{\circ}$ or $ 90^{\circ}$.

True, but there is a purely synthetical solution.

Here is an idea of my solution: let I be an incentre of ABC, and IF be perpendicular to AC (F belongs to AC). If F=E, we have an answer 60, if not - then FEIK is cyclic and answer is 90. And it is not difficult to check this two variants.

Let $ K,H$ be the incenter of triangle $ ADC, ABC.AK\cap HE = O$. Put $ AB = AC = x, BC = y$ $ \angle AKH = \angle HEK = 45^o$ then $ HK^2 = HO.HE$ But ${ \frac {HO}{HE} = \frac {AH}{AH + AE}, \frac {HK^2}{HC^2} = \frac {HD^2}(HD + DC)^2}$ We get $ \frac {HC^2.HD^2}{(HD + DC)^2} = \frac {HE^2.AH}{AH + AE}$ $ \Rightarrow \frac {HD^2}{(HD + DC)^2} = \frac {HE^2}{HC^2}.\frac {AH}{AH + AE} = \frac {AE^2}{AB^2}.\frac {AH}{AH + AE}$ $ \Rightarrow \frac {HD}{(HD + DC)^2} = \frac {AE^2}{AB^2}.\frac {AH}{HD(AH + AE)}$ $ \Rightarrow \frac {HD}{(HD + DC)^2} = \frac {AE^2}{AB^2}.\frac {AB}{BD(AH + AE)} = \frac {AE^2}{AB.BD.(AH + AE)}$ $ \Rightarrow AE^2.(HD + DC)^2 = HD.AB.BD.(AH + AE)$ $ \Rightarrow \frac {x^4}{(x + y)^2}.(\frac {1/2y.AD}{1/2y + x} + 1/2y)^2 = \frac {1/2y.AD}{1/2y + x}.x.1/2y.(\frac {x.AD}{1/2y + x} + \frac {x^2}{x + y})$ $ \Rightarrow \frac {x^2}{(x + y)^2}.(\frac {1/2AD}{1/2y + x} + 1/2)^2 = \frac {1/2AD}{1/2y + x}.1/2.(\frac {AD}{1/2y + x} + \frac {x}{x + y})$ Then set $ x = ky$ and note that $ AD = \sqrt {k^2 -1/4}y$ we can easy find $ k$. $ \Rightarrow \angle A = 60^o$ or $ 90^o$. PS: Bad proof but easy if we calculate

Solution by the proposers can be found here.

Peter wrote: True, but there is a purely synthetical solution. orl wrote: Solution by the proposers can be found here. Like that

Let $ D$ be the midpoint of $ BC$; $ I$ be the incenter of $ \triangle ABC$. $ DK$ meets $ AC$ at point $ T$.Let $ AB=AC=a$;$ BC=2$. It means $ CD=DB=1$ Firstly, triangle $ IEK$ and triangle $ CDT$ are similar (since $ \angle IED=\angle EDC = 45^o$). Thus $ \frac{IE}{IK}=\frac{CD}{CT}$. We have: $ \frac{IK}{ID}=\frac{IC}{ID+1} \rightarrow \frac{1}{IK}=\frac{ID+1}{ID.IC}$ $ \frac{IE}{EC}=\frac{IB}{2}\rightarrow IE=\frac{EC.IC}{2}$ (notice that $ IB=IC)$. Thus $ \frac{IE}{IK}=(1+\frac{1}{ID})\frac{EC}{2}$ In triangle $ ABC$ : $ \frac{EC}{BC}=\frac{EC}{2}=\frac{AC}{AB+BC}=\frac{a}{a+2}$. But $ \frac{ID}{DC}=\frac{AD}{AC+DC}=\frac{AD}{a+1}$ Hence $ \frac{IE}{IK}=\frac{(AD+a+1)a}{AD(a+2)}$. On other hand; $ \frac{CD}{CT}=\frac{AD}{AT}=\frac{1+AD}{a}$. It implies \[ \frac{(AD+a+1)a}{AD(a+2}=\frac{1+AD}{a}\] \[ \leftrightarrow AD.a^2+a^3+a^2=a.AD+a.AD^2+2.AD+2.AD^2\] \[ \leftrightarrow AD.a^2=a.AD-a+2AD+a^2-2\] (since $ AD^2=a^2-1$) \[ \leftrightarrow AD(a^2-a-2)=a^2-a-2\] Thus either $ AD=1$ or $ a^2-a-2=0$. $ AD=1$ implies $ \angle A =90^o$ $ a^2-a-2=0$ implies $ a=2$ thus $ A=60 ^o$. We have done!.

Simple solution by trigonometry calculation Let $ \alpha = \measuredangle DAC$, and consider triangle $ %Error. "traingel" is a bad command. EJC$, where $ J$ is the incenter of $ \triangle ACD$. According to the assumption, this triangle has angles $ \frac{3\alpha}{2}, 45^\circ - \frac{\alpha}{2}, 135^\circ - \alpha$. From Sine law in $ \triangle BEC$, we have \[ \frac{EC}{\sin 45^\circ - \alpha/2} = \frac{2R \sin 2 \alpha}{\sin {45^\circ - 3\alpha/2}}.\] From Sine law in $ \triangle AJC$, we have \[ \frac{JC}{\alpha/2} = \frac{2R \sin {90^\circ - \alpha}}{\sin 135^\circ}.\] Finally, in triangle $ \triangle EJC$, we have \[ \frac{EC}{\sin{135^{\circ} - \alpha}} = \frac{JC}{\sin {3 \alpha / 2}}.\] After simplification, we have following identity \[ \frac{\sin {45^\circ + \alpha}}{\sin \alpha} \cdot \frac{\sin {45^\circ + 3\alpha/2}}{\sin {3 \alpha/2}} = \sqrt{2} \cdot \frac{\sin {45^\circ - \alpha/2}}{\sin {\alpha/2}}.\] Now, set $ x = \cot \frac{\alpha}{2}$ and using formulas for double and triple angle, it follows \[ \left(\frac{x^2-1}{2x}+1\right)\left(\frac{3x-x^3}{1-3x^2}+1\right)=2(x-1).\] This equation is equivalent to $ (x-1) (x^2-4x+1) (x^2-2x-1)=0$. It follows that $ \cot 2x = \ctg \alpha = \sqrt{3}$ or $ \cot 2x = \ctg \alpha = 1$. This brings us to $ \alpha = 45^\circ$ or $ \alpha = 30^\circ$, and finally $ \measuredangle CAB = 90^\circ$ or $ \measuredangle CAB = 60^\circ$.

And thus Hojoo manages to squeeze one more IMO problem in Good job guys!

Yeah, I too have solved it by trigonometry but it is not an elegant one after all. I tused the cotangent rule in $ BEC$. That is for a trinagle $ ABC$ divided in the ratio $ m: n$ internally by a cevian making angle $ \theta$ , along with the base angles being $ \alpha$ and $ \beta$ respectively, $ (m+n)\cot \theta = m\cot \alpha + n \cot \beta$. THis laong with the angle bisector theorem got me trhough the sum. TOok some time still. Yes hojoo lee is having his 8th problem in to the IMO {i think so , probably more}.

Agr_94_Math wrote: Yes hojoo lee is having his 8th problem in to the IMO {i think so , probably more}. I think it is problem number 9.

Valentin Vornicu wrote: And thus Hojoo manages to squeeze one more IMO problem in Good job guys! I really enjoyed our cowork. Whenever Jan and Peter created breakthroughs on our problems, I was very happy!

Congratulations on 8(9?) th problem on IMO!

$ I \equiv AD \cap BE$ is incenter of isosceles $ \triangle ABC.$ Reflect $ E \in AC$ in $ CI$ into $ F \in BC,$ $ EF \perp CI.$ $ \angle IDK = 45^\circ.$ Therefore $ \angle BEK = \angle IEK = \angle IFK = 45^\circ$ $ \Longleftrightarrow$ $ F \in \odot(KDI)$ $ \Longleftrightarrow$ $ F \equiv D$ or $ KIDF$ is cyclic. The 1st possibility leads to equilateral $ \triangle ABC,$ $ \angle CAB = 60^\circ.$ The 2nd possibility $ \Longleftrightarrow$ $ \angle IKF = \angle IDF = 90^\circ$ $ \Longleftrightarrow$ $ K \in EF.$ From $ \triangle BEC,$ $ 180^\circ = \angle EBC + \angle BCE + \angle CEK + \angle KEB =$ $ = \frac {_1}{^2} \angle ABC + \angle BCA + 90^\circ - \frac {_1}{^2} \angle BCA + 45^\circ$ $ \Longleftrightarrow$ $ \angle ABC + \angle BCA = 90^\circ$ $ \Longleftrightarrow$ $ \angle CAB = 90^\circ.$

Quote: Let $ \triangle ABC$ with $ AB = AC$ . The angle bisectors of $ \angle CAB$ and $ \angle ABC$ meet $ BC$ and $ CA$ at $ D$ and $ E$ respectively. Let $ K$ be the incentre of $ \triangle ADC$. Suppose that $ \angle B E K = 45^\circ$ . Find all possible values of $ \angle C AB$ . Proof. Denote $ \boxed {\ A = 4x\ }$ . Thus, $ \left\|\begin{array}{cc} \triangle IDC\ : & \frac {KI}{KC} = \frac {DI}{DC} = \tan \widehat {DCI} = \tan \left(45^{\circ} - x\right) \\ \\ \triangle IEC\ : & \frac {KI}{KC} = \frac {\sin\widehat {ECI}}{\sin\widehat {EIC}}\cdot \frac {\sin\widehat {KEI}}{\sin\widehat {KEC}} = \frac {\sin\left(45^{\circ} - x\right)}{\sin \left(90^{\circ} - 2x\right)}\cdot\frac {\sin 45^{\circ}}{\sin 3x}\end{array}\right\|\ \implies$ $ \cos\left(45^{\circ} - x\right)\sin 45^{\circ} =$ $ \sin 3x\cos 2x\ \Longleftrightarrow\ \cos x + \sin x =$ $ 2\sin 3x\cos 2x = \sin 5x + \sin x\ \Longleftrightarrow$ $ \cos x = \cos\left(90^{\circ} - 5x\right)\ \Longleftrightarrow\ \left(\ x = 90^{\circ} - 5x\ \right)\ \ \vee\ \ \left(\ x + 90^{\circ} - 5x = 0\ \right)$ $ \stackrel{(A=4x)}{\ \Longleftrightarrow\ }\boxed {\ A\ \in\ \{\ 60^{\circ}\ ,\ 90^{\circ}\ \}\ }$ .

Virgil Nicula wrote: Quote: Let $ \triangle ABC$ with $ AB = AC$ . The angle bisectors of $ \angle CAB$ and $ \angle ABC$ meet $ BC$ and $ CA$ at $ D$ and $ E$ respectively. Let $ K$ be the incentre of $ \triangle ADC$. Suppose that $ \angle B E K = 45^\circ$ . Find all possible values of $ \angle C AB$ . Proof. Denote $ \boxed {\ A = 4x\ }$ . Thus, $ \left\|\begin{array}{cc} \triangle IDC\ : & \frac {KI}{KC} = \frac {DI}{DC} = \tan \widehat {DCI} = \tan \left(45^{\circ} - x\right) \\ \\ \triangle IEC\ : & \frac {KI}{KC} = \frac {\sin\widehat {ECI}}{\sin\widehat {EIC}}\cdot \frac {\sin\widehat {KEI}}{\sin\widehat {KEC}} = \frac {\sin\left(45^{\circ} - x\right)}{\sin \left(90^{\circ} - 2x\right)}\cdot\frac {\sin 45^{\circ}}{\sin 3x}\end{array}\right\|\ \implies$ I am sorry. But why should $ \frac {KI}{KC} = \frac {DI}{DC} = \frac {\sin\widehat {ECI}}{\sin \widehat {EIC}}\cdot \frac {\sin\widehat {KEI}}{\sin\widehat {KEC}}$.Sorry for the inconvenience.

There is much much simpler geometrical solution. I will ensure that it is correct and post then ... It looks correct. $ I$ is incentre. $ E'$ is symmetrical of $ E$ with respect of $ CK$. Then $ \angle IE'K = \angle IEK = 45$, but $ \angle IDK = 45$, so $ DIKE'$ is concyclic. It means that $ \angle EIK = \angle E'IK = \angle IDC = 45$. From this we get $ \angle A = 90$. In $ D = E'$ case we get $ \angle A = 60$. It is simple to check that both satisfy given condition. ... just noticed yetti wrote the same

Virgil Nicula wrote: An easy extension. Let $ \triangle ABC$ with the incentre $ I$ . Denote $ D\in AI\cap BC$ , $ E\in BI\cap AC$ and the incentre $ K$ of $ \triangle ADC$ . Prove that $ \widehat {B E K}\equiv\widehat {ADK}\ \implies\ A\ \in\ \{B\ ,\ 2C\ \}$ . Proof. $ \left\|\begin{array}{cc} \triangle IDC\ : & \frac {KI}{KC} = \frac {DI}{DC} = \frac {\sin \widehat {DCI}}{\sin\widehat {DIC}}=\frac {\sin \frac C2}{\sin\left(90^{\circ}-\frac B2\right)}\\ \\ \triangle IEC\ : &{ \frac {KI}{KC} = \frac {\sin\widehat {ECI}}{\sin\widehat {EIC}}\cdot \frac {\sin\widehat {KEI}}{\sin\widehat {KEC}} = \frac {\sin\frac C2}{\sin\frac {B+C}{2}}\cdot\frac {\sin \frac {A+2B}{4}}{\sin \frac {3A}{4}}}\end{array}\right\|\ \implies$ $ \cos\frac B2\sin\left(\frac A4+\frac B2\right)=\cos \frac A2\sin\frac {3A}{4}\ \Longleftrightarrow\ \cos\left(\frac A4+B\right)+\sin\frac A4=$ $ \sin\frac {5A}{4}+\sin\frac A4 \Longleftrightarrow$ $ \cos\left(\frac A4+B\right)=$ $ \sin\frac {5A}{4}$ $ \Longleftrightarrow$ $ \left(\frac A4+B=\frac {5A}{4}\right)\ \ \vee\ \ \left(\frac A4+B+\frac {5A}{4}=180^{\circ}\right)$ $ \Longleftrightarrow$ $ \boxed {\ A\ \in\ \left\{\ B\ ,\ 2C\ \right\}\ }$ .

Let $I$ be the incenter of triangle $ABC$ and by Quadrilateral Ceva on $CDIE$ $$\frac{\sin45\cdot\sin\angle ECI\cdot\sin45\cdot\sin\angle CID}{\sin\angle KEC\cdot\sin\angle ECI\cdot\sin45\cdot\sin\angle CIE}=1.$$Let $\angle ABC=2\alpha$ and note that $\angle CID=90-\alpha,$ $\angle KEC=135-3\alpha,$ and $\angle CIE=2\alpha.$ Hence, \begin{align*}-\cos135&=\sin45\\&=(\sin(135-3\alpha)\cdot\sin(2\alpha))/(\cos\alpha)\\&=2\sin\alpha\cdot\sin(135-3\alpha)\\&=\cos(4\alpha-135)-\cos(135-2\alpha).\end{align*}Therefore, $$\cos(135-2\alpha)=2\cos(135-2\alpha)\cos(2\alpha)$$and $\angle ABC=60,45$ or $\angle CAB=60,90.$ $\square$

Note that ∠IEK = ∠IDK = 45 so reflecting E across CK would be nice. Let I be incenter of ABC and E' be reflection of E across CK. Note that CI is angle bisector so E' lies on BC. now we have IKDE' is cyclic and we have 2 cases : 1 - E' is D : It means BE is perpendicular to AC so ABC is regular triangle and ∠CAB = 60. 2 - E' isn't D : ∠EKI = ∠E'KI = ∠E'DI = 90 and ∠IEK = 45 so ∠CIE = 45 so ∠CAB = 90. we're Done.

:crying: why trig We claim that the answer is $60^\circ,90^\circ.$ First, note it's true when $\angle CAB=60^\circ$ or $90^\circ.$ Let $F$ be the incenter of $\triangle ABC$ Now, we reflect $E$ over $CK$ to $E'.$ Note that $\angle FDK=45^\circ=\angle FE'K=45^\circ.$ Given that $D\neq E',$ $FDE'K$ is cyclic. Thus, $\angle FKE'=90^\circ.$ This means that $\angle KEC=180^\circ-\angle AEK=135^\circ-\angle ECB-\angle EBC=90^\circ-\angle ECK$ which implies $45^\circ=\angle FCD+\angle FBC$ so $\angle BAC=90^\circ.$ If $D=E'$ then $EC=CD$ which implies that $AC=BC$ so $\angle BAC=60^\circ$

oops totally didnt already see solution The answers are $\angle CAB=\boxed{60^{\circ}}$ and $\angle CAB=\boxed{90^{\circ}}$. Let $I$ be the incenter of $ABC$. Set $\angle CAB=4x$. Angle chasing gives $\angle EIK=90-2x$, $\angle ECK=45-x$, $\angle CEK=3x$, $\angle KEI=45$, $\angle KCD=45-x$, $\angle KID=45+x$. Now \[\frac{\sin(45-x)}{\sin(3x)}\div \frac{\sin(90-2x)}{\sin(45)}=\frac{EK}{KC}\div \frac{EK}{IK}=\frac{IK}{KC}=\frac{ID}{CD}=\frac{\sin(45-x)}{\sin(45+x)}.\]So we obtain \[\frac{1}{2}(\cos(x)-\cos(90+x))=\sin(45)\sin(45+x)=\sin(3x)\sin(90-2x)=\frac{1}{2}(\cos(5x-90)-\cos(90+x))\]which means that \[2\sin(3x-45)\sin(2x-45)=\cos(x)-\cos(5x-90)=0\implies x=15, \frac{45}{2}\implies \angle CAB=\boxed{60^{\circ},90^{\circ}}.\]We are done. $\blacksquare$

bad asdf imagien not covering up the sol so abd Let $I$ be the incenter of $\triangle ABC$. There is a homothety at $C$ mapping $I$ to $K$ to $C$, so $I, K, C$ are collinear. We proceed using trig bash. Since $K$ is the incenter of $\triangle ADC$, we know by the Angle Bisector Theorem that $$\frac{IK}{KC} = \frac{ID}{DC} = \frac{\sin\angle ICD}{\sin\angle DIC} = \frac{\sin \left(45^{\circ}-\frac A4\right)}{\sin \left(45^{\circ}+\frac A4\right)}.$$Also, by LoS on $\triangle IKE$ and $\triangle EKC$, we have \begin{align*}\frac{IK}{\sin 45^{\circ}} &= \frac{EK}{\sin\left(90^{\circ} - \frac A2\right)} \\ \frac{KC}{\sin\left(\frac{3A}{4}\right)} &= \frac{EK}{\sin\left(45^{\circ} - \frac A4\right)},\end{align*}so\begin{align*}\frac{IK}{KC} = \frac{\frac{\sin 45^{\circ}}{\sin\left(90^{\circ} - \frac A2\right)}}{\frac{\sin\left(\frac{3A}{4}\right)}{\sin\left(45^{\circ} - \frac A4\right)}} &= \frac{\sin\left(45^{\circ} - \frac A4\right)}{\sin\left(45^{\circ} + \frac A4\right)}\\ \sin45^{\circ} \cdot \sin\left(45^{\circ} + \frac A4\right) &= \sin\left(\frac{3A}{4}\right)\cdot\sin\left(90^{\circ} - \frac A2\right) \\ \cos\left(\frac A4\right) - \cos\left(90^{\circ} + \frac A4\right) &= \cos\left(\frac{5A}{4}-90^{\circ}\right) - \cos\left(90^{\circ} + \frac A4\right) \\ \cos\left(\frac A4\right) &= \cos \left(\frac{5A}{4} - 90^{\circ}\right).\end{align*}It's not hard to verify that $\boxed{A = 60^{\circ}, 90^{\circ}}$ are the solutions. We are done. $\square$ hahahaa i didnt format any of the latex so its jsut a clomp of text

Let $\angle DAK=\alpha$, so $\angle BAC=4\alpha$. Let $I$ be the incenter of $\triangle ABC$. We have $\angle AIE=\angle DIC=\alpha+45$, so $\angle KEC=2\alpha+\alpha+45-\alpha=3\alpha$. Thus, $$\frac{KC}{IC}=\frac{\sin(90-2\alpha)}{\sin45}\cdot\frac{\sin 3\alpha}{\sin(45-\alpha)}.$$However, by triangle $\triangle IDC$, we have $$\frac{KC}{IC}=\frac{DC}{ID}=\tan(\alpha+45)$$Hence, $$\frac{\sin(90-2\alpha)}{\sin45}\cdot\frac{\sin3\alpha}{\sin(45-\alpha)}=\tan(\alpha+45)$$$$\sin(90-2\alpha)\sin3\alpha=\sin45\sin(45-\alpha)\tan(\alpha+45).$$However, $\sin(45-\alpha)\tan(45+\alpha)=\sin(45+\alpha)$, so we have $$\sin(90-2\alpha)\sin3\alpha=\sin45\sin(45+\alpha).$$Multiplying by 2 this becomes $$\cos(90-5\alpha)-\cos(90+\alpha)=\cos\alpha-\cos(\alpha+90)$$$$\cos(90-5\alpha)=\cos\alpha.$$This gives $\alpha=\pi/8$ and $\alpha=\pi/12$ as solutions, which correspond to answers of $60$ and $90$ degrees. These clearly work (e.g. by coordbashing), so we are done. edit: didn't see EK perp CK wow

We claim the answer is 60 or 90 degrees. I got a hint to reflect E over K to get E', and let I the incenter of ABC. Suppose $D\ne E'$. Then IDK=1/2IDC=45=IEK=IE'K, so IDEK is cyclic. Now, let CEE'=CE'E=x. Then $$180-2x=ECB=180-CBE-BEC=180-(ACB/2)-45-x=135-x-(90-x)=45\implies x=\frac{135}2\implies BAC=180-2ACB=180-2(45)=90.$$On the other hand, if $D=E'$ it follows that CD=CE; since CD is a tangent to the incircle, E lying on AC implies CE is a tangent too; in particular, BE perp. AC whence ABC is equilateral, so BAC=60 degrees. trig bashing CAN be done but will be omitted here due to bashing sobad

Let $I$ be the incenter of $\triangle ABC$. Since both $CI$ and $KC$ bisect $\angle ACB$, we have that $K$, $I$, and $C$ are collinear. Let $\angle ACI=a$. Notice that since $\triangle ABC$ is isosceles, we have that \[\angle AIE=\angle BID=\angle DIC=90-\angle DCI=90-a,\]and that \[\angle AEI=180-(\angle AIE+\angle AEI)=180-((90-2a)+(90-a))=3a,\]which in turn gives us that $\angle KEC=180-(45+3a)=135-3a$. Now using Law of Sines on $\triangle IEK$ and $\triangle KEC$, we get that \[\frac{IK}{KC}=\frac{EK*\frac{\sin 45}{\sin 2a}}{EK*\frac{\sin (135-3a)}{\sin a}}=\frac{\sin 45\sin a}{\sin 2a \sin (135-3a)}.\]However, by Angle Bisector Theorem and Law of Sines on $\triangle IDC$, we get that \[\frac{IK}{KC}=\frac{ID}{DC}=\frac{\sin a}{\cos a},\]and setting the two equations equal, we get that \[\frac{\sin 45\sin a}{\sin 2a \sin (135-3a)}=\frac{\sin a}{\cos a} \iff \sin 45\cos a=\sin 2a\sin (135-3a).\]By Product-to-Sum, this gives us that \[\frac{1}{2}(\sin (a+45)+\sin(45-a))=\frac{1}{2}(\cos(135-5a)-\cos(135-a)),\]however, since $-cos(135-a)=\sin(45-a)$ and $\cos(135-5a)=-\sin(45-5a)$, this gives us \[\sin(a+45)+\sin(45-5a)=0,\]which is \[2\sin(45-2a)\cos(3a)=0,\]by Sum-to-Product. Since $a<45$, we find that either $a=30$ or $a=\frac{45}{2}$, which gives us that the only possible values for $\angle BAC$ are $60$ and $90$ degrees, finishing the problem.

Let $I$ of the incenter of $\triangle ABC$. Consider the reflection $E'$ of $E$ over $CI$, which we know will land on $BC$. If $D = E'$, the angle bisector $AI$ is an altitude, forcing $\triangle ABC$ to be equilateral, or $\angle CAB = \boxed{60}$. Otherwise, the angle equality $\angle IDK = \angle IE'K = 45$ tells us $IDE'K$ is cyclic, which induces \[\angle IKE' = \angle IKE = 90 \implies EKE' \text{ collinear} \implies \angle IKA = \angle EKA = 45\]\[\implies \triangle EKA \cong \triangle IKA \implies \angle AIE = \angle AEI,\] from which it's easy to find $\angle B = \angle C = 45$, so $\angle CAB = \boxed{90}$. $\blacksquare$

Motivation Let's make a few observations: By the angle condition $\angle{BEK}=45^{\circ}=\angle{EDK}$, we know that $BEKD$ is cyclic. $K$ being the incenter of $\triangle{ADC}$ is a key point. $K$ is the intersection of lines $\overline{EK}, \overline{CK}, \overline{BK}$ in $\triangle{BEC}$. $\Rightarrow{}$ Can we use Ceva's theorem on $\triangle{BEC}$? Solution Let $\angle{B}=\angle{C}=\theta$. Then, $\angle{CEK}=135^{\circ}-\frac{3\theta}{2}$. Also, notice that $\angle{EFK}=\theta$. By Trig Ceva on $\triangle{BEC}$, we have \begin{align*} \frac{\sin(45^{\circ})}{\sin(\angle{CEK})}*\frac{\sin(\angle{CBK})}{\sin(\angle{EBK})}*\frac{\sin(\angle{ECK})}{\sin(\angle{BCK})}&=1 \\ \frac{\sin(45^{\circ})}{\sin(135^{\circ}-\frac{3\theta}{2})}*\frac{DK}{EK}&=1 \text{ (by $BEKD$ being concyclic)}\\ \frac{\sin(45^{\circ})}{\sin(135^{\circ}-\frac{3\theta}{2})}*\frac{\sin(90^{\circ}-\frac{\theta}{2})}{\sin(\theta)}&=1 \text{ (by the Law of Sines on $\triangle{DFK}$ and $\triangle {EKF}$)} \\ \frac{1}{\sin(\frac{3\theta}{2})+\cos(\frac{3\theta}{2})}*\frac{1}{2\sin(\frac{\theta}{2})}&=1 \\ 2\sin(\frac{\theta}{2})*\sin(\frac{3\theta}{2})+2\sin(\frac{\theta}{2})*\cos(\frac{3\theta}{2})&=1 \\ -\cos(2\theta)+\cos(\theta)+\sin(2\theta)-\sin(\theta)&=1 \text{ (by product to sum)} \\ -2\cos^2(\theta)+\cos(\theta)+2\sin(\theta)\cos(\theta)-\sin(\theta)&=0 \\ (2\cos(\theta)-1)(-\cos(\theta)+\sin(\theta))&=0 \\ \end{align*} Note that $\theta < 90^{\circ}$, thus either $\cos(\theta)=\frac{1}{2}$ or $\sin(\theta)=\cos(\theta)$. We can conclude that $\boxed{\theta=60^{\circ}, 90^{\circ}}$.

bad formatting because copied from my solution on mathdash consider the reflection of D over the angle bisector of C, denote this point as F. then we have IFK = IDK = 45, thus we say IDFK is concyclic. if F \neq E, we get 90 = IFE = IKE, so since IEK = 45, we have 45 = KIE = CIE. letting ACI = x, we get that IEC = 180 - EBC - ECB = 180 - x - 2x = 180 - 3x. now we have ICE + CIE + CEI = 180, so 180 - 3x + x + 45 = 180, so 2x = 45, implying CAB = 90, which obviously works. now if F = E, we are forced that ABC is equilateral. it is trivial to then see that this case actually works, since IEK = IDK = 45 since they are reflections over IK. so CAB can eb 60 as well.

I bashed this on paper so I'll just post the highlights here (I know that this wouldn't be a valid writeup but I'm not going to type up the same bash I did by hand). Swap the names of points $B$ and $D$. Note that $\angle ABK=45^{\circ}$ so quadrilateral $DBKE$ is cyclic. We employ barycentric coordinates wrt $ABC$. $D=(0,2,-1)$, $E=(2a:0:b)$, $K=(a:b:c)$, $B=(0,1,0)$. The circle has equation $-a^2yz-b^2xz-c^2xy+(ux+vy+wz)(x+y+z)=0$. Plug in $B$ to get $v=0$, plug in $D$ to get $w=2a^2$, plug in $E$ to get \[u=\frac{b^3-2a^2b-ab^2}{2a+b}.\]Plug in $K$, immediately we can cancel out a factor of $a+b+c$. Eventually we get \[(c-a)(c^2+2ac-a^2)=b(c-a)(c+a).\]So $c=a$ or we can square both sides and cancel stuff out to get $a=\sqrt3 c$, so the answers are $\boxed{60^{\circ},90^{\circ}}$. These both work because the steps are reversible to show that $K$ lies on $(BDE)$, which implies the angle condition. $\blacksquare$

Let $\angle ABE = \angle EBC = \angle ACK = \angle KCB = \alpha$. Now $\angle ABC = 180 - 4\alpha$ $\Rightarrow$ $\angle AEB = 3\alpha$ $\Rightarrow$ $\angle KEC = 180 - 3\alpha - 45 = 135 - 3\alpha$. Also it is obvious that $I \in KC$, since CK is the angle bisector of $\angle BCA$. Now by law of sines in $\triangle IEK$ we get that $IK = \frac{EK.\sin 45}{\sin 2\alpha}$. Again by law of sines in $\triangle KEC$ we get that $KC = \frac{EK.\sin (135 - 3\alpha)}{\sin \alpha}$ $\Rightarrow$ now we get that $\frac{IK}{KC} = \frac{\sin 45.\sin \alpha}{\sin 2\alpha.\sin (135 - 3\alpha)}$. Also since K is incenter in $\triangle ADC$, DK is an angle bisector in $\triangle IDC$ $\Rightarrow$ $\frac{IK}{KC} = \frac{ID}{DC} = tg \alpha = \frac{\sin \alpha}{\cos \alpha}$ $\Rightarrow$ we get that $\frac{IK}{KC} = \frac{\sin 45.\sin \alpha}{\sin 2\alpha.\sin (135 - 3\alpha)} = \frac{\sin \alpha}{\cos \alpha}$ $\Rightarrow$ $\sin 45.\cos \alpha = \sin 2\alpha.\sin (135 - 3\alpha)$, which using trig formulas gets down to $\frac{1}{2}.(\sin(45 + \alpha) + \sin(45 - \alpha)) = \frac{1}{2}.(\cos(5\alpha - 135) - \cos(135 - \alpha))$. Since $\cos(135 - \alpha) = \sin(\alpha - 45) = - \sin(45 - \alpha)$, we get that $\sin(45 + \alpha) = \cos(5\alpha - 135) = \cos(135 - 5\alpha) = \sin(5\alpha - 45) = - \sin(45 - 5\alpha)$ $\Rightarrow$ $\sin(45 + \alpha) + \sin(45 - 5\alpha) = 0$ $\Rightarrow$ $\sin(45 + \alpha) + \sin(45 - 5\alpha) = 2\sin(45 - 2\alpha)\cos 3\alpha = 0$ $\Rightarrow$ $3\alpha = 90$ or $2\alpha = 45$ $\Rightarrow$ $\alpha = 30^{\circ}$; $\frac{45}{2}^{\circ}$ and since $\angle CAB = 180 - 4\alpha$, then $\angle CAB = 180 - 4.30 = 60^{\circ}$ or $\angle CAB = 180 - 4.\frac{45}{2} = 90^{\circ}$ $\Rightarrow$ $\angle CAB = 60^{\circ}; 90^{\circ}$.

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Due to symmetry, $\angle A = 60^{\circ}$ works fine. From hereon, assume $\angle A \neq 60^{\circ}$. Let $I$ be the incenter of $\triangle ABC$ and let $D'$ be the reflection of $D$ across line $CI$. Then, $D'$ lies on $\overline{AC}$ and we have $\angle ID'K = \angle IDK = 45^{\circ} = \angle IEK$. Since, by assumption, $D' \neq E$ (this is crucial), we must have \[CD' \cdot CE = CK \cdot CI \implies \tfrac{CD}{CK} = \tfrac{CI}{CE},\]so by SAS similarity we have $\triangle CDK \sim \triangle CIE$. This implies $\angle CIE = 45^{\circ}$, so $\angle A = 90^{\circ}$.