From the Angle Bisector theorem we have
|AD| = bc/(a+b) and |BD| = ac/(a+b).
Let the extension of CD intersect the circumcircle of
ΔABC at point E. Since CD bisects ∠ACD,
∠ACD = ∠DCB = ∠ECB.
Inscribed angles subtended by the same chord are
equal. Therefore,
∠CAD = ∠CAB = ∠CEB.
Therefore,
ΔCAD ~ ΔCEB and
|CA| |CE| |CD|+|DE|
------ = ------ = -----------
|CD| |CB| |CB|
or
|CB||CA| = |CD|2 + |CD||DE|
From the Intersecting Chords theorem we have
|CD||DE| = |AD||BD|. Therefore,
|CB||CA| = |CD|2 + |AD||BD|
or
|CD|2 = |CB||CA| - |AD||BD|
or
|CD|2 = ab - [bc/(a+b)][ac/(a+b)]
or
|CD|2 = ab[ 1-{c/(a+b)}2 ]
QED
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