mestavk wrote: In a triangle $ABC$ with a right angle at $A$, $H$ is the foot of the altitude from $A$. Prove that the sum of the inradii of the triangles $ABC$, $ABH$, and $AHC$ is equal to $AH$ We have $ \begin{cases} r_{ABC}=\frac{1}{2}(AB+AC-BC) \\ r_{ABH}=\frac{1}{2}(AH+BH-AB) \\ r_{AHC}=\frac{1}{2}(AH+HC-AC) \end{cases} $ Now $\boxed{r_{ABC}+r_{ABH}+r_{AHC} = AH }$

Note that the area of a triangle is equal to the product of its inradius and its semiperimeter; i.e., $|\triangle ABC| = rs$.* Letting $BC = a$, $CA = b$, $AB = c$, and $AH = h$, we then have \[ r = \frac{\frac{1}{2}ah}{\frac{1}{2}(a+b+c)} = \frac{ah}{a+b+c}. \] If $r_B$ is the inradius of $\triangle HBA$ and $r_C$ is the inradius of $\triangle HAC$, then by similarity of triangles, we also have \[ \frac{r_B}{r} = \frac{AB}{CB} = \frac{c}{a}, \quad \frac{r_C}{r} = \frac{CA}{CB} = \frac{b}{a}. \] Consequently, \[ r + r_B + r_C = r\left(1 + \frac{b}{a} + \frac{c}{a}\right) = r \cdot \frac{a+b+c}{a} = \frac{ah (a+b+c)}{(a+b+c)a} = h. \] *Note. The proof of this fact is easily demonstrated by drawing the line segments joining the incenter $I$ to the three vertices $A, B, C$, then observing that the altitudes from $I$ to each side are all equal to the inradius $r$. Thus $|\triangle ABC| = |\triangle AIB| + |\triangle BIC| + |\triangle CIA| = \frac{rc}{2} + \frac{ra}{2} + \frac{rb}{2} = rs$, as claimed.