An easy but nice geometry problem

[ins-]

Prove that for a right-angled triangle $ABC$ ($\angle{C}=90^o$) with exradii $r_a$, $r_b$, $r_c$ the following equality is correct: $(r_a-r_c)^2+(r_b-r_c)^2=(r_a+r_b)^2$.

[Sergic Primazon]

ins- в сообщении #1116634 писал(а):

Prove that for a right-angled triangle $ABC$ ($\angle{C}=90^o$) with exradii $r_a$, $r_b$, $r_c$ the following equality is correct: $(r_a-r_c)^2+(r_b-r_c)^2=(r_a+r_b)^2$.

$$R_b=CL=AP, R_a=CQ=PB, R_a+R_b=AB$$

$$R_c-R_b=ML=LA+AN=CK+PB=CZ+ZB=CB$$

$$R_c-R_a=XQ=QB+BN=CZ+AP=CK+AK=AC$$

$$AC^2+BC^2=AB^2 )))$$

$

[ins-]

The reverse statement is also correct. It is true that a triangle is right-angled iff $r_ar_b=rr_c$. http://forumgeom.fau.edu/FG2006volume6/FG200639.pdf here are interesting statements on this subject.