Equal Incircles Theorem, Angela Drei's Proof
Let C be a point. Assume points Mi, I = 1, 2, ..., N (N > 3) lie on a line not through C. Assume further that the incircles of triangles M1CM2, M2CM3, ..., MN-1CMN all have equal radii. Then the same is true of triangles M1CM3, M2CM4, ..., MN-2CMN, and also of triangles M1CM4, M2CM5, ..., MN-3CMN, 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.
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.
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.
Given a triangle ABC, D lie on AB, r1 and r2 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 - 2r1/h)(1 - 2r2/h) = 1 - 2r/h.
In particular, if r1 = r2, then (1 - 2r1/h)² = 1 - 2r/h.
Apply Lemma 1 to triangles ACD, BCD, and ABC and observe that ∠ADC/2 = 90° - ∠BDC/2 so that
Assume points Mi, I = 1, 2, ..., N (N > 3) lie on side AB of ΔABC,
(1 - 2ρ/h)N-1 = 1 - 2r/h.
Similar to the proof of Lemma 2, apply Lemma 1 to each of the triangles MiCMi+1 and multiply the identities. The products of the tangents due to pairs of supplementary angles at points Mj,
Proof of the Equal Incircle Theorem
This is a direct consequence of Lemma 2 applied to pairs of adjacent triangles Mi-1CMi and MiCMi+1.
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