Cathetus? What does that mean?

nsun48 wrote: Point $C_{1}$ of hypothenuse $AC$ of a right-angled triangle $ABC$ is such that $BC = CC_{1}$. Point $C_{2}$ on cathetus $AB$ is such that $AC_{2} = AC_{1}$; point $A_{2}$ is defined similarly. Find angle $AMC$, where $M$ is the midpoint of $A_{2}C_{2}$. Easy problem. Since $CC_1=CB$ and $CA_2=CA_1$ then $\triangle{CC_1A_2}=\triangle{CBA_1}$ and $\angle{CC_1A_2}=\angle{CBA_1}$. Triangles $AC_1C_2$ and $AA_1B$ are isosceles and similar, so $\angle{C_2C_1A_2}=180^0-(\angle{AC_1C_2}+\angle{CC_1A_2})=180^0-(\angle{ABA_1}+\angle{CBA_1})=90^0$. Analogically $\angle{C_2A_1A_2}=90^0$. Thus, points $C_2, C_1, A_1, A_2$ lie on the circle with diameter $A_2C_2$, $MC_2=MC_1=MA_2=MA_1$. Since also $AC_1=AC_2, CA_1=CA_2$ then by the symmetry we get, that $AM$ and $CM$ are the bisectors of the angles $BAC$ and $BCA$ respectively. Thus, $\angle{AMC}=135^0$.

The problem can be solved this way as well: knowing that, in a right-angled triangle the following equality holds: $a+c=2(R+r)$, where $a,c,R,r$ are the legs, circumradius (i.e. half of hypotenuse) and the inradius respectively, by construction of $A_2, C_2$ we easily get $BA_2=BC_2=2r$ and, trivially, $M$ is the incenter of $\triangle ABC$, done. Best regards, sunken rock

$BC_2=BA_2=2(s-b)$ and so $M$ is the incenter and trivial angle chasing. (its perp. projections have distance $s-b$ to $B$)

This is my solution Denote $I$ is the incenter of triangle $ABC$. Because of $AC_1=AC_1; AB=AA_1; CA_1=CA_1; CB=CC_1$ so $I$ is on the perpendicular bisectors of $C_1C_2, A_1A_2, C_1B, A_1B$. That shows: $I$ is the circumcenter of triangle $BC_1C_1, BA_2A_1$. And because the circles $(BC_2C_1)$ and $(BA_1A_2)$ have a same center and have a common point $B$, so they are the same circle. That means $I$ is the circumcenter of $(BA_2C_2)$, So $I$ is the midpoint of $A_2C_2$ or $I \equiv M$. We have $\angle{AMC}=135^0$.

Let $\angle {A_2MC}=x,\angle {C_2MA}=y$ We've $MA_2^2=\frac {(a+c-b)^2}{2}$ Using sine rule in $A_2MC$ and $C_2MA$ respectively we've $\frac {a}{b-c}=cot x,\frac {c}{b-a}=cot y$ Now using $b^2=a^2+c^2$ we've $x+y=45^0$....so done.

$BA_{1}C_{1}C_{2}$ and $BA_{2}A_{1}C_{1}$ are isosceles trapezoids so $BA_{2}A_{1}C_{1}C_{2}$ is cyclic with circumcenter $M$ cause $\angle A_{2}BC_{2}=90$ now $A,M$ are both on the bisector of $BA_{1}$ which is also the bisector of $\angle CAB$ similarly $M$ is on the bisector of $\angle BCA$ so $M$ is incenter of $ABC$ and $\angle AMC=135$