Let the bisector of /B intersect AD at point E.
Let the line through E, parallel to line AC,
intersect line BC at point F.
/DFE = /C = /B/2 = /DBE
Therefore, right triangles EDF and EDB are
congruent. Thus, |DF| = |BD|.
From the similar triangles ADC and EDF and
how the bisector of an angle splits the
opposite side,
|FC| |AE| |AB|
------ = ------ = ------
|DF| |ED| |BD|
Therefore, |FC| = |AB|.
Thus,
|AB| + |BD| = |FC| + |DF|
= |DF| + |FC|
= |DC|.
Note: See Harry's post for an
alternate solution.
|
