Why statement is given like this? Problem is equal to this: $ABC$ is a triangle with right angle $A$. Prove that $r_{A}=r_{B}+r_{C}+r$. And it's easy evaluate all of lengths.

So yeah i realized some of the stuff in my solution were unnecessary but i only discovered after finishing the problem but a solution is a solution and there is way too many points in this problem so i will not state them. I think its easy to see that the problem is basically about the incircle and the excircles of the triangle $ABC$ so let $R$ be the inradius of $\triangle ABC$, $r_{a}$ the radius with center $I_{a}$, $r_{b}$ the radius with center $I_{b}$ and $r_{c}$ the radius with center $I_{c}$. We know that the internal and the external bissector of an angle are perpendicular so $\angle I_{c}AI_{a}=\angle I_{c}CI_{a}=90^\circ$ so $I_{c}ACI_{a}$ is cyclic therefore $\angle I_{a}AC=\angle CI_{c}I_{a}=45^\circ$ because of the square $AFIE$ but notice $\angle IBI_{c}=90^\circ$ so $\angle BII{_c}=45^\circ$ and now since $I_{c}I{_b}$ and $BI$ bissect externally and internally at $\angle B$ we have $\angle IBK=\angle IBF=\angle BI_{c}V$ which means $\triangle BIK \cong \triangle I_{c}BV (A.S.A)$ Let $L$ be the intersection of $I_{c}P$ and $DT$ such that $DL // I_{c}I_{a}$ but we already know that $I_{a}D // I_{c}P$ so $I_{c}I_{a}DL$ is a paralellogram so $I_{c}L=I_{a}D$ and $\angle I_{a}I_{c}L=\angle LDI_{a}=90^\circ + \angle IBK$ but $\angle I_{a}DP=90^\circ$ $\implies$ $\angle PDL=\angle IBK$. Notice that because of the right angles we have $ST // PD$ so $\angle STL=\angle PDL$ , $I_{c}VST$ is a rectangle and $I_{c}BTL$ is a paralellogram so $I_{c}V=BK=ST$ and $IB=I_{c}B=LT$ therefore $\triangle STL \cong KBI (S.A.S)$ $\implies$ $SL=BV=IK=R$. We have that $PSI_{b}U$ is a rectangle so $PS=UI_{b}=r_{b}$, for the same reason $I_{c}P=I_{c}V=r_{c}$ and finally since $I_{c}I_{a}LD$ is a paralellogram $I_{c}L=I_{a}D=I_{c}P + PS + SL$ $\implies$ $r_{c} + r_{b} + R= r_{a}$ as we wanted .

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