Let I be the incenter of triangle ABC and Gamma the circle with diameter AI.
Since angles AABI and AACI are right angles,
AB and AC lie on Gamma. Let B' be the orthogonal projection of I onto side AC.
Since angle AB'I is a right angle, B' lies on Gamma.
/ABAAC = 180 - /ABIAC = /BIAC = /IBC + /ICB
= (/B + /C)/2 = 90 - /A/2 = 90 - /IAB'
= /AIB'
Since chords ABAC and AB' subtend equal inscribed angles in the same circle,
|ABAC| = |AB'| = s - a.
Similar arguments give |BCBA| = s - b and |CACB| = s - c. Therefore,
|ABAC| + |BCBA| + |CACB| = (s - a) + (s - b) + (s - c) = s.
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