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Attachments:
Sharygin_P3.pdf (162kb)

Ok so i'll show only for one case because the other case is quite similiar. Proof: Let $F$ be the foot of altitude from $E$ onto $AC$ and let $O$ be the circumcentre of $\Delta ADE$. Easy to see that $FA$ is tangent to $\odot(ADE)$ since $\angle FAE= \angle ADE$(alternate segment theorem) and so $\angle OAF=90^\circ$. Its known by the problem that $OA=OE=OF \implies \angle EOF =\angle EFO=90^\circ-\angle AFO=\angle AOF \implies OF $ is an angle bisector in $\Delta AOE$ and since $\Delta AOE$ is iscoseles $\implies OF$ is a perpendicular bisector in $\Delta AOE \implies FA=FE$ since $OF$ will also be an perpendicular bisector in $\Delta FAE$ which implies that it is iscoseles and so we have $FA=FE \implies \Delta FAE$ is iscoseles and right. Now notice that $\Delta FAE \sim \Delta CAB$ since $\angle AFE=\angle ACB =90^\circ \implies EF \parallel BC$ .By this we have $\Delta CAB$ is iscoseles and right $\implies$ angles of the triangle are $45^\circ,45^\circ,90^\circ$

Not commenting! Sharygin 2020 CR P3 wrote: Let $ABC$ be a triangle with $\angle C=90^\circ$, and $D$ be a point outside $ABC$, such that $\angle ADC=\angle BAC$. The segments $CD$ and $AB$ meet at point $E$. It is known that the distance from $E$ to $AC$ is equal to the circumradius of triangle $ADE$. Find the angles of triangle $ABC$. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(8cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -9.96, xmax = 9.76, ymin = -5.19, ymax = 5.69; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw((-2.24,0.35)--(3.88,0.37), linewidth(0.4) + wrwrwr); draw((0.66,3.39)--(-2.24,0.35), linewidth(0.4) + wrwrwr); draw((3.88,0.37)--(0.66,3.39), linewidth(0.4) + wrwrwr); draw((0.66,3.39)--(-1.1,-2.09), linewidth(0.4) + wrwrwr); draw((-2.24,0.35)--(-1.1,-2.09), linewidth(0.4) + wrwrwr); draw(circle((-1.2737723864558959,-0.6848772625244757), 1.415827231573573), linewidth(0.4) + wrwrwr); draw((-0.3143292420841493,0.3562930416925355)--(-1.3222044460558817,1.3121029255138346), linewidth(0.4) + wrwrwr); draw((-0.3143292420841493,0.3562930416925355)--(-1.2737723864558959,-0.6848772625244757), linewidth(0.4) + wrwrwr); draw((-1.2737723864558959,-0.6848772625244757)--(-2.24,0.35), linewidth(0.4) + wrwrwr); /* dots and labels */ dot((-2.24,0.35),dotstyle); label("$A$", (-2.6,0.47), NE * labelscalefactor); dot((3.88,0.37),dotstyle); label("$B$", (3.96,0.57), NE * labelscalefactor); dot((0.66,3.39),dotstyle); label("$C$", (0.74,3.59), NE * labelscalefactor); dot((-1.1,-2.09),dotstyle); label("$D$", (-1.18,-2.55), NE * labelscalefactor); dot((-0.3143292420841493,0.3562930416925355),linewidth(4pt) + dotstyle); label("$E$", (-0.12,0.43), NE * labelscalefactor); dot((-1.2737723864558959,-0.6848772625244757),linewidth(4pt) + dotstyle); label("$O$", (-1.28,-1.09), NE * labelscalefactor); dot((-1.3222044460558817,1.3121029255138346),dotstyle); label("$K$", (-1.38,1.53), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Let $EK\perp AC$ where $K\in AC$, and let $O$ be the center of $\odot(ADE)$. So we are given that $EK=EO$. From the given angle condition it is clear that $\odot(ADE)$ is tangent to $AC$ at $A$. So, $OA\perp AC$. So, as $EK=EO$ and $\angle EKA=\angle OKA=90^\circ$. Hence, $OAKE$ is a square. So, $\angle EKA=\angle EOA=90^\circ\implies \angle ADE=\angle CAB=45^\circ$. So, $\angle CAB=45^\circ$ and $\angle ABC=45^\circ$. $\blacksquare$

$\textbf{Diagram : }$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -10.6, xmax = 10.6, ymin = -6.38, ymax = 6.38; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw((-7.96,4.18)--(-7.78,-2.02), linewidth(1) + wrwrwr); draw((-7.78,-2.02)--(-0.5,-1.74), linewidth(1) + wrwrwr); draw((-0.5,-1.74)--(-7.96,4.18), linewidth(1) + wrwrwr); draw((-2.68,4.14)--(-7.78,-2.02), linewidth(1) + wrwrwr); draw((-7.96,4.18)--(-2.68,4.14), linewidth(1) + wrwrwr); draw((-4.753552772583052,1.6354735138996865)--(-7.883488515178252,1.5446044116953097), linewidth(1) + wrwrwr); /* dots and labels */ dot((-7.96,4.18),dotstyle); label("$A$", (-7.88,4.38), NE * labelscalefactor); dot((-0.5,-1.74),dotstyle); label("$B$", (-0.42,-1.54), NE * labelscalefactor); dot((-7.78,-2.02),dotstyle); label("$C$", (-8.14,-2.32), NE * labelscalefactor); dot((-2.68,4.14),dotstyle); label("$D$", (-2.58,4.24), NE * labelscalefactor); dot((-4.753552772583052,1.6354735138996865),linewidth(4pt) + dotstyle); label("$E$", (-4.6,1.5), NE * labelscalefactor); dot((-7.883488515178252,1.5446044116953097),linewidth(4pt) + dotstyle); label("$F$", (-7.8,1.7), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] $$\textbf{Solution : }$$Let the perpendicular from $E$ to $AC$ meet $AC$ at $F$. Let $\angle ADC = \angle BAC = x$. $\linebreak$ We know that in $\triangle ADE$, $\frac{AE}{\sin x}=2R$, where $R$ is the circumradius of $\triangle ADE$ (By the Extended Law of Sines - see Theorem 3.1 Page 43 - Euclidean Geometry in Mathematical Olympiads by Evan Chen). But, $\sin x = \tfrac{AF}{AE}.$ So, $AE^2 = 2EF^2 \implies EF = AF$, by Pythagorean Theorem, and also $\angle FAE = \angle FEA = 45^{\circ}. $ Hence, the angles of the $\triangle ABC$ are $90^{\circ}, 45^{\circ}, 45^{\circ}. $

@all above I guess you'all forgot that there's something called Law of sines. We (I ) would show that the angles of triangle are $45^\circ, 45^\circ, 90^\circ$. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; usepackage("amsmath"); size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.5) + fontsize(8); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -3.3612712130003843, xmax = 6.222210626675132, ymin = -1.9196767762920808, ymax = 4.6231986989681015; /* image dimensions */ draw((0,3)--(0,0)--(3,0)--cycle, linewidth(0.4)); draw(arc((3.6114281985907724,1.7338266825210866),0.23879107573942315,160.67924058443643,205.6454045119977)--(3.6114281985907724,1.7338266825210866)--cycle, linewidth(0.4)); draw(arc((0,3),0.23879107573942315,-90,-45)--(0,3)--cycle, linewidth(0.4)); /* draw figures */ draw((0,3)--(0,0), linewidth(0.4)); draw((0,0)--(3,0), linewidth(0.4)); draw((3,0)--(0,3), linewidth(0.4)); draw((0,3)--(3.6114281985907724,1.7338266825210866), linewidth(0.4)); draw((3.6114281985907724,1.7338266825210866)--(0,0), linewidth(0.4)); draw((2.0268976572205464,0.9731023427794537)--(0,0.9731023427794537), linewidth(0.4)); label("$\theta$",(0.05344117007336689,2.8959432511196104),SE*labelscalefactor); label("$\theta$",(3.1656848572105156,1.9168998405879774),SE*labelscalefactor); label("$R$",(0.8255323149641685,1.3119624487147732),SE*labelscalefactor); /* dots and labels */ dot((0,3),linewidth(1pt) + dotstyle); label("$A$", (-0.04207526022240238,3.071056706661854), NE * labelscalefactor); dot((3,0),linewidth(1pt) + dotstyle); label("$B$", (3.062208724390099,-0.041186980475288855), NE * labelscalefactor); dot((0,0),linewidth(1pt) + dotstyle); label("$C$", (-0.1694305006167614,-0.06506608804923113), NE * labelscalefactor); dot((3.6114281985907724,1.7338266825210866),linewidth(1pt) + dotstyle); label("$D$", (3.6591864137386567,1.686068467373202), NE * labelscalefactor); dot((2.0268976572205464,0.9731023427794537),linewidth(1pt) + dotstyle); label("$E$", (1.9796891810380473,0.8025414871373381), NE * labelscalefactor); dot((0,0.9731023427794537),linewidth(1pt) + dotstyle); label("$F$", (-0.1694305006167614,0.8900982149084598), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let the circumradius of $\triangle ADE$ be equal to $R$. Then, it follows that $EF = R$. Now, by using the Law of sines in $\triangle ADE$, we obtain $$AE=2R\sin\theta.$$Now, as $AFE$ is a right triangle, we obtain $$AE=\frac{EF}{\sin\theta}=\frac R{\sin\theta}.$$Equating both the equations, yields $$2R\sin\theta = \frac R{\sin\theta}\implies \sin^2\theta = \frac12,$$and thus, $\theta=45^\circ$, as desired. $\blacksquare$

NJOY wrote: @all above I guess you'all forgot that there's something called Law of sines. Yup I forgot.

here is my solution

Attachments:
p3.pdf (369kb)