Consider a triangle ABC. Let the angle bisector of angle A intersect side BC at a point D between B and C.
The angle bisector theorem states that the ratio of the length of the line segment BD to the length of segment DC is equal to the ratio of the length of side AB to the length of side AC.
BD/DC= AB/AC
Prove it.
GEOMETRY SOLUTION
Let E be the intersection of line AC and
the line through B and parallel to AD.
/AEB = /CAB = /BAD = /ABE
==> |EA| = |AB|
Clearly triangles ACD and ECB are similar.
Therefore,
|AC| |EC| |EA|+|AC| |AB|+|AC|
------ = ------ = ----------- = -----------
|CD| |CB| |CD|+|DB| |CD|+|BD|
|AB| |AC|
==> ------ = ------
|BD| |CD|
TRIGONOMETRY SOLUTION
Rule of sines applied to triangle ABD:
|AB| |BD|
----------- = -----------
sin(/ADB) sin(/BAD)
Rule of sines applied to triangle ACD:
|AC| |AC| |CD| |CD|
----------- = ----------- = ----------- = ----------
sin(/ADB) sin(/ADC) sin(/CAD) sin(/BAD)
|AB| |AC|
==> ------ = ------
|BD| |CD|
QED
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Posted by Bractals
on 2015-11-07 11:53:19 |
Solutions
