$ABC$ is a right triangle such that $\angle B= 90^{\circ}$ and $BD$ is the altitude to $AC$. Given that:
$I$ is the incenter of $\triangle ABC$, $I_1$ is the incenter of $\triangle ABD$ and $I_2$ is the incenter of $\triangle CBD$.
We have to prove that: $$IB=I_1I_2$$ using pure geometry and preferably without much algebra. I can prove it relatively straightforwadly using the using the formula $rs=\Delta$ but the algebra is a bit tedious. Does anyone have a neat geometric solution? :)
To prove: Let ABC be a right triangle with right angle at B and D the projection of B onto AC. Let I be the incenter of incircle $\Sigma$ of ABC, $I_1$ the incenter of incircle $\Sigma_1$ of ABD and $I_2$ the incenter of incircle $\Sigma_2$ of BCD. Then $|BI|=|I_1I_2|$.
Proof: Let E be the projection of $I_1$ onto BD and F the projection of $I_2$ onto BD. Let $r={a+c-b\over 2}$ be the radius of $\Sigma$. Clearly triangles ABD and BCD are similar to ABC. So the radius of $\Sigma_1$ is $r_1=cr/b$ and the radius of $\Sigma_2$ is $r_2=ar/b$. Obviously $|DE|=r_1$ and $|DF|=r_2$, so $|EF|=|r_1-r_2|$. Thus the distance $|I_1I_2|$ is the length of the diagonal of the rectangle with sides $r_1+r_2$ and $|r_1-r_2|$; $$|I_1I_2|=\sqrt{(r_1+r_2)^2+(r_1-r_2)^2}=\sqrt2\cdot\sqrt{r_1^2+r_2^2}=\sqrt2\,r\cdot\sqrt{{c^2+a^2\over b^2}}=\sqrt2\,r=|BI|$$