Denote the triangle sides a = BC, b = CA, c = AB. Let (O) be the circumcircle of the right angle triangle $\triangle ABC$ centered at the midpoint O of its hypotenuse c = AB. Let R, S, T be the tangency points of the circles $K_1, K_2, K_3$ with the line AB. In an inversion with the center A and positive power $r_A^2 = AC^2 = b^2$ ($r_A$ being the inversion circle radius), the line AB is carried into itself, the circle (O) is carried into the altitude line CD and the altitude line CD into the circle (O). This implies that the circle $K_3$ intersecting the inversion circle (A) is carried into itself, but this is possible only if the circle $K_3$ is perpendicular to the inversion circle (A). It follows that the tangency point T of the circle $K_3$ is the intersection of the inversion circle $(A, r_A = b)$ with the line AB. Similarly, in an inversion with the center B and positive power $r_B^2 = BC^2 = a^2$ ($r_B$ being the inversion circle radius), the line AB is carried into itself, the circle (O) is carried into the altitude line CD and the altitude line CD into the circle (O). This implies that the circle $K_2$ intersecting the inversion circle (B) is carried into itself, but this is possible only if the circle $K_2$ is perpendicular to the inversion circle (B). It follows that the tangency point S of the circle $K_2$ is the intersection of the inversion circle $(B, r_B = a)$ with the line AB. The distance between the tangency points S, T is the equal to ST = AT - AS = AT - (AB - BS) = AC - (AB - BC) = a + b - c. The radius r of the incircle $K_1$ of the right angle triangle $\triangle ABC$ is equal to $r = \frac{|\triangle ABC|}{s} = \frac{ab}{a + b + c} = \frac{a + b - c}{2} = s - c$ where $|\triangle ABC|$ and s are the area and semiperimeter of the triangle $\triangle ABC$, for example, because of an obvious identity $(a + b + c) (a + b - c) = a^2 + 2ab + b^2 - c^2 = 2ab$ or just because the angle $\angle C = 90^\circ$ is right. Therefore, ST = 2r. Let R' be the midpoint of ST. Then $AR' = AT - \frac{ST}{2} = b - r = b - \frac{a + b - c}{2} = \frac{c + b - a}{2} = s - a = AR$ Therefore, the points $R' \equiv R$ are identical and the midpoint of the segment ST is the tangency point R of the incircle $K_1$ with the triangle side c = AB. It follows that the normals to the hypotenuse AB at the tangency points S, T of the circles $K_2, K_3$ are tangent to the incircle $K_1$. Radii $r_2, r_3$ of the circles $K_2, K_3$ are now easily calculated: $r_2 = SD = BS - BD = a - \frac{a^2}{c}$ $r_3 = TD = AT - AD = b - \frac{b^2}{c}$ Denote $I, I_2, I_3$ the centers of the circles $K_1, K_2, K_3$. The line $I_2I_3$ cuts the midline RI of the trapezoid $STI_3I_2$ at the distance from the point R equal to $\frac{SI_2 + TI_3}{2} = \frac{r_2 + r_3}{2} = \frac{a + b}{2} - \frac{a^2 + b^2}{2c} = \frac{a + b - c}{2} = r = RI$ As a result, the centers $I_2, I, I_3$ are collinear (in fact, I is the midpoint of the segment $I_2I_3$). The common center line $I_2I_3$ and the common external tangent AB of the circles $K_1, K_2, K_3$ meet at their common external homothety center $H \equiv I_2I_3 \cap AB$ and the other common external tangent of the circles $K_2, K_3$ from the common homothety center H is a tangent to the circle $K_1$ as well.

It directly follows from Sawayama-Thebault's Theorem(http://en.wikipedia.org/wiki/Th%C3%A9bault%27s_theorem)

An outline of my solution (I wrote it twice, but the system restarted unexpectedly before sending it): Let $M, N, P$ be contact points of the 3 circles with $AB$, and $I, Q, R$ their centers. As shown before and well known, $BN=BC, AP=AC$; also $\Delta CNP$ has the circumcenter $I$ and $\angle NCP=45^\circ$, hence $IM=MN=MP$. With $QN=ND, PR=PD$, we get $QN+PR=ND+PD=NP=2MI$, so $MI$ is the midline of the trapezoid $QNPR$ and, $Q-I-R$ are collinear, hence the reflection of $AB$ about $QR$ is the second common tangent of the three circles. Best regards, sunken rock