Al menos el triangulo rectangulo isosceles cumple con el problema, es decir, <A=<B=45 y <C=90 [Moderator says: Please write in english in this forum. Translation: At least, the isosceles right triangle fulfills the problem condition, that is, <A=<B=45 and <C=90]

Not sure if its typed correctly, circle through $CXY$ as $\Gamma$ constitutes to a harder problem... $H_1, H_2$ be orthocentres of them respectively. $CH_1=CH_2, CX=CY, H_1X=H_2Y$ as both perpendicular to $AB$. Therefore, a flip across the angle bisector at $C$ swaps the triangles i.e. $H_1H_2 \parallel XY$. So, $H_1H_2XY$ is a rectangle. This means $XY \perp YH_2 \perp AB \implies XY \parallel AB \implies CAB$ is isosceles and $C \in YH_1, XH_2 \implies H_1, H_2$ lie on $BC, AC$ respectively meaning $BC \perp AC$. Therefore, $CAB$ is an isosceles right triangle, with $C = \dfrac{\pi}{2}$.

Let $H_1$ be the orthocentre of $\Delta ABX$ and $H_2$ the orthocentre of $\Delta AYB$. $AH_1\perp BC$ and $BH_2\perp CA$. Let $BC\cap AH_1=D$ and $BH_2\cap AC=Y$ As $CX=CH_1$ and $AMXD$ is a cyclic quad.($M$ is the foot of the altitude from $X$ on $BC$.), $\angle DXH_1=90^{\circ}-B$ So, $\angle BCH_1=180^{\circ}-2(90^{\circ}-B)=2B$ $\therefore r=H_1C=\frac{CD}{\cos 2B}=\frac{b\cos C}{\cos 2B}$ By doing the same in $\Delta ABY$ we have $r=\frac{a \cos C}{\cos 2A}$ Therefore, $\frac{b\cos C}{\cos 2B}=\frac{a \cos C}{\cos 2A}\implies a\cos 2B=b\cos 2A$ Form the lemma mentioned above, we have $a=b$ and using this, it is easy to prove that $ABC$ must be a right angled isosceles triangle.

IDMasterz wrote: Not sure if its typed correctly, circle through $CXY$ as $\Gamma$ constitutes to a harder problem... $H_1, H_2$ be orthocentres of them respectively. $CH_1=CH_2, CX=CY, [color=\#FF0000]H_1X=H_2Y[/color]$ as both perpendicular to $AB$. Therefore, a flip across the angle bisector at $C$ swaps the triangles i.e. $H_1H_2 \parallel XY$. So, $H_1H_2XY$ is a rectangle. This means $XY \perp YH_2 \perp AB \implies XY \parallel AB \implies CAB$ is isosceles and $C \in YH_1, XH_2 \implies H_1, H_2$ lie on $BC, AC$ respectively meaning $BC \perp AC$. Therefore, $CAB$ is an isosceles right triangle, with $C = \dfrac{\pi}{2}$. $H_1X$ is not equal $H_2Y$

mathandyou wrote: IDMasterz wrote: Not sure if its typed correctly, circle through $CXY$ as $\Gamma$ constitutes to a harder problem... $H_1, H_2$ be orthocentres of them respectively. $CH_1=CH_2, CX=CY, [color=\#FF0000]H_1X=H_2Y[/color]$ as both perpendicular to $AB$. Therefore, a flip across the angle bisector at $C$ swaps the triangles i.e. $H_1H_2 \parallel XY$. So, $H_1H_2XY$ is a rectangle. This means $XY \perp YH_2 \perp AB \implies XY \parallel AB \implies CAB$ is isosceles and $C \in YH_1, XH_2 \implies H_1, H_2$ lie on $BC, AC$ respectively meaning $BC \perp AC$. Therefore, $CAB$ is an isosceles right triangle, with $C = \dfrac{\pi}{2}$. $H_1X$ is not equal $H_2Y$ True, I falsely assumed they had to be inside the triangle lol. Anyway, if they are outside, its still true $H_1Y=H_2X$. With this, you can use some angle chasing and extremal arguments to still jump to the same conclusion. Will post solution later

let $D,E$ be orthocentre of triangle $ABX,ABY$ respectively than , $DX,EY$ are perpendicular to $AB$ and hence parallel to each other thus $XYED$ is a isosceles trapezoid. and thus $XY=DE$ as $CY=CX$ thus $\angle CYX=\angle CXY = 90-C/2$ now by sine law in triangle $CYX$ we have , $\frac {CX.sin C}{cos C/2} = XY$ similarly angle chasing we get $\angle CED=\angle CDE = C/2$ hence sine law in triangle $CED$ gives $\frac {CE.sin C}{sin C/2} = DE=XY$ thus we get $sin C/2=cos C/2$ as $CE=CX$ as both lie on same circle. thus $\angle C=90$ and now as $AC,AD$ both are perpendicular to $BC$ gives that $A,C,D$ must be collinear. and hence $\angle ADX=\angle YCX = \angle YCX/2 = C/2=45$ since angle subtended at center of circle is twice the angle subtended at circumference and hence in triangle $ADK$ where $K$ is feet of perpendicular from $X$ on $AB$ that ,$\angle A = 45$ similarly we get $\angle B = 45$ thus the given triangle is right angled isosceles triangle. and we are done