Question 444866
{{{drawing(300,300,0,10,0,10,


triangle(2,2,8,2,8,9),
locate(2,2,A),
locate(8,2,C),
locate(8,9,B),
line(2,2,8,4.5),
locate(8,4.5,D)
)
}}}


If *[tex \angle BAC = \theta], you know that


*[tex \LARGE \sin{\frac{\theta}{2}} = \frac{CD}{AD}]
*[tex \LARGE \cos{\frac{\theta}{2}} = \frac{BD}{AD}]
*[tex \LARGE \tan{\frac{\theta}{2}} = \frac{CD}{BD}]


Also, you know that AD is an angle bisector, so the following ratio holds:

*[tex \LARGE \frac{AB}{BD} = \frac{AC}{CD}]


Try combining these to derive the half-angle formulas.