Let D, E, and F be the intersections with side AB
of the altitude, bisector, and median respectively.
The order of the points being A, D, E, F, and B.
Let x = ∠ACD = ∠DCE = ∠ECF = ∠FCB.
Using the law of sines on triangles ACF and FCB we get
sin(90-x) sin(∠CAF) |CF|
----------- = ----------- = ------
sin(3x) sin(∠ACF) |AF|
|CF| sin(∠CBF) sin(90-3x)
= ------ = ----------- = ------------
|FB| sin(∠FCB) sin(x)
or
sin(x)cos(x) = sin(3x)cos(3x)
or
sin(2x) = sin(6x)
Either 2x = 6x or 2x + 6x = 180°.
Obviously, 4x = 90° and x = 22.5°. Therefore,
|AC|
------ = tan(∠ABC)
|BC|
= tan(22.5°) = √2 - 1
~= 0.414214
|
