Equal Incircles Theorem, Angela Drei's Proof
The Equal Incircle Theorem is a generalization of a sangaku problem:
Let C be a point. Assume points M_{i}, I = 1, 2, ..., N (N > 3) lie on a line not through C. Assume further that the incircles of triangles M_{1}CM_{2}, M_{2}CM_{3}, ..., M_{N1}CM_{N} all have equal radii. Then the same is true of triangles M_{1}CM_{3}, M_{2}CM_{4}, ..., M_{N2}CM_{N}, and also of triangles M_{1}CM_{4}, M_{2}CM_{5}, ..., M_{N3}CM_{N}, and so on.
The proof below is by Angela Drei, an Italian mathematics teacher. The proof first appeared in an Italian magazine Archimede. It depends on two lemmas.
Lemma 1
In any ΔABC,
1  2r/h = tan(α/2)tan(β/2),
where r is the inradius, h the altitude to AB, α = ∠BAC, β = ∠ABC.
Proof
Let a, b, c be the lengths sides BC, AC, AB and p = (a + b + c)/2 the semiperimeter of ΔABC. As we know,
tan²(α/2) = (p  b)(p  c) / p(p  a) and tan²(β/2) = (p  a)(p  c) / p(p  b)
Taking the product gives
tan²(α/2)tan²(β/2) = (p  c)² / p².
Next, letting S be the area of ΔABC,
(p  c) / p  = 1  c/p  
= 1  ch·r / pr·h  
= 1  2Sr / Sh  
= 1  2r / h, 
which proves the lemma.
Lemma 2
Given a triangle ABC, D lie on AB, r_{1} and r_{2} are the radii of the circles inscribed in ACD and BCD, r is the inradius, h is the length of the altitude to the side AB. Then the following relation holds:
(1  2r_{1}/h)(1  2r_{2}/h) = 1  2r/h.
In particular, if r_{1} = r_{2}, then (1  2r_{1}/h)² = 1  2r/h.
Proof
Apply Lemma 1 to triangles ACD, BCD, and ABC and observe that ∠ADC/2 = 90°  ∠BDC/2 so that
Lemma 2'
Assume points M_{i}, I = 1, 2, ..., N (N > 3) lie on side AB of ΔABC,
(1  2ρ/h)^{N1} = 1  2r/h.
Proof
Similar to the proof of Lemma 2, apply Lemma 1 to each of the triangles M_{i}CM_{i+1} and multiply the identities. The products of the tangents due to pairs of supplementary angles at points M_{j},
Proof of the Equal Incircle Theorem
This is a direct consequence of Lemma 2 applied to pairs of adjacent triangles M_{i1}CM_{i} and M_{i}CM_{i+1}.
Sangaku

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