How to prove $x^2+y^2=z^2$ for $x,y,z$ are inradius of $\triangle{ADC},\triangle{ADB},\triangle{ABC}$ In a $\triangle{ABC}$, $AD$ is altitude through $A$ ;$x,y,z$ are inradius of $\triangle{ADC},\triangle{ADB},\triangle{ABC}$. How to prove $x^2+y^2=z^2$.I have no idea how to do this, can this be proved with simple geometry?
 A: I agree with @chenbai: this property is false.
Here is a proof: Let us take a fixed coordinate system with $H$ is the origin, the $x$-axis is the base of the triangle, the $y$-axis is the altitude  (see figure below).
Let us fix as well the two "small" incircles $(C_1)$ and $(C_2)$, with resp. radii, say $5$ and $12$ for example, tangent to the $x$-axis and the $y$ axis as shown on the figure. Thus, the radius of the "large" incircle should be $13$. We will see that it's not the case in general.
The essential fact is that there is an infinite number of triangles $ABC$ having these "small" incircles for triangles $AHB$ and $AHC$. 
It suffices, for each position $(0,h)$ of $A$ on the $y$ axis, to take the second external tangent to $(C_1)$ and to $(C_2)$, $B$ and $C$ being thus obtained as the intersection of these tangents with the $x$-axis.
If $A(O,h)$ moves on the $y$-axis, the line bissectors issued from $B$ and $C$ move themselves ; as a consequence, their intersection point $I$, the center of the "large" incircle, moves, and this evolution has a component on the $y$ axis; thus the radius of the "large" cannot be constant. A contradiction.
In particular, let us consider the extreme cases where the ordinate $h$ of $A$ is arbitrarily large: the line bissectors will be arbitrarily close to the straight lines making a $\pi/4$ angle with the $x$-axis. The limit line bissectors have equations $y=x+10$ and $y=-x+26$ ; they intersect at $I(8,18)$, showing that the radius of the "large" incircle is $18$ instead of $13$... 

