A paper triangle has its vertices cut off. Each cut is along a straight line
parallel to the side opposite the vertex and tangent to the triangle's incircle.

Prove that the triangle's inradius is equal to the sum of inradii of the
three triangles cut off.

Let ΔABC be the paper triangle with a, b, and c the lengths of the sides
opposite vertices A, B, and C respectively. Let r and s be the inradius and
semiperimeter. Let h_{a}, h_{b}, and h_{c} be the altitudes through vertices A, B,
and C respectively.

Let B_{1}C_{1} be the cut for vertex A with B_{1} on AB and C_{1} on AC,
A_{2}C_{2} the cut for vertex B with A_{2} on AB and C_{2} on BC, and
A_{3}B_{3} the cut for vertex C with A_{3} on AC and B_{3} on BC.
Let r_{a}, r_{b}, and r_{c} be the inradii of ΔAB_{1}C_{1},
ΔA_{2}BC_{2}, and ΔA_{3}B_{3}C respectively.