Let c = |AB|, b = |AC|, d = |AD|, x = |BD|, a-x = |CD| and r the common inradius.
Applying the Law of Cosines to triangles ABC and ABD,
b2 = c2 + a2 - 2ac cos(B)
and
d2 = c2 + x2 - 2ac cos(B)
Combining these and eliminating cos(B), we get
a(c2 + x2 - d2) = x(c2 + a2 - b2) (1)
Letting ha be the altitude on BC we get
Area(ABD) = ½xha = ½(c + d + x)r
Area(ACD) = ½(a - x)ha = ½(b + d + a - x)r
Combining these and eliminating ha and r, we get
x = a(c + d)/(b + c + 2d) (2)
Plugging x from (2) into (1) we get
(d + b)(d2 - s[s - a]) = 0
where s = (a + b + c)/2.
Therefore, |AD| = √(s[s - a]).
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