△ A B C contains four inscribed circles. Seqments J I , K L , and M H are parallel to the sides of △ A B C which they do not intersect. Compare the radius of circle X to the sum of the radii of circles Y , Z , and W .
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The triangles A B C , A L K , M B H and J I C are all similar, and so we deduce that Z = h a h a − 2 X X W = h b h b − 2 X X Y = h c h c − 2 X X where h a , h b , h c are the lengths of the altitudes of the triangle A B C from the vertices A , B , C respectively. This is because the length of the altitude from A in the triangle A L K is h a − 2 X , and so on. Thus Y + Z + W = 3 X − 2 X 2 ( h a − 1 + h b − 1 + h c − 1 ) = 3 X − Δ X 2 ( a + b + c ) = 3 X − Δ 2 X 2 s = 3 X − 2 X = X where Δ , s are the area and semiperimeter of the triangle A B C .
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Since J I ∥ A B , H M ∥ C A and K L ∥ C B , △ A B C ∼ △ J I C ∼ △ M B H ∼ △ A L K ⟹ A B M B = X W , A B J I = X Y , A B A L = X Z ( 1 ) Observe, △ A B C ∼ △ J I C ∼ △ M F L Also, △ J I C and △ M F L have the same exradius ( X ) , so we have, △ J I C ≅ △ M F L ⟹ J I = L M ( 2 ) Using ( 1 ) and ( 2 ) , A B M B + A B J I + A B A L ⟹ A B M B + L M + A L ∴ X = W + Y + Z = X W + X Y + X Z = X W + Y + Z