Can you prove or disprove the following claim:
Claim. Given any cyclic quadrilateral $ABCD$. Consider a triangle defined by any triple of the vertices of quadrilateral. Define exradius of the triangle's excircle that touches side of the quadrilateral as outer exradius, and exradius of the triangle's excircle that touches a diagonal of the quadrilateral as inner exradius. Then, no matter how one triangulates a quadrilateral $ABCD$, the sum of outer exradii of triangles is constant.
Let us denote exradii of triangle $\triangle ABC$ - $r_1,r_2,r_3$ , exradii of triangle $\triangle ACD$ - $r_4,r_5,r_6$ , exradii of triangle $\triangle ABD$ - $r_7,r_8,r_9$ and exradii of triangle $\triangle BCD$ - $r_{10},r_{11},r_{12}$ , where $r_3,r_6,r_9,r_{12}$ are inner exradii. Using the Japanese theorem for cyclic polygons and identity $r_a+r_b+r_c=4R+r$ ,where $r_a,r_b,r_c$ are exradii, $R$ - circumradius and $r$ - inradius one can show that: $r_1+r_2+r_3+r_4+r_5+r_6=r_7+r_8+r_9+r_{10}+r_{11}+r_{12}$ . Now, in order to finish the proof I need to show that: $r_3+r_6=r_9+r_{12}$ and I don't know how to do that. Any hints are welcomed.