PRINCIPIA · THEOREM

Angle bisector length t² = ab − mn

Dependencies: Angle bisector ⇔ equidistant from sides (the bisector + its property), Inscribed angles on same arc are equal; angle in a semicircle is right (inscribed angles subtending the same arc are equal), AA similarity (AA similarity), Intersecting chords: PA·PB = PC·PD (the equal-product theorem for two intersecting chords inside a circle).

Statement

In △ABC\triangle ABC, let ADAD be the interior angle bisector of ∠A\angle A, with DD on BCBC (by Angle bisector divides opposite side in ratio of adjacent sides, DD divides BCBC into BD=mBD = m and DC=nDC = n). Denote the two adjacent sides

b  =  AC,c  =  AB,b \;=\; AC,\qquad c \;=\; AB,

and write the length of the interior bisector segment as t=ADt = AD. Then

  t2  =  bc  −  mn.  \boxed{\;t^2 \;=\; bc \;-\; mn.\;}

That is, the square of the interior bisector length = the product of the two adjacent sides − the product of the two segments it cuts off on the opposite side. Combined with Angle bisector divides opposite side in ratio of adjacent sides (mn=cb\dfrac{m}{n} = \dfrac{c}{b}), this lets one compute tt directly from a,b,ca, b, c.

△ABC inscribed in its circumscribed circle ⊙O, AD the interior bisector of ∠A, extended to meet ⊙O at E; t² = bc − mn

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