✔ 最佳答案
In △ABD , by sine formula , BD / sin(a + b)
= AD / sin∠ABD
= AD / sin∠ACD ..... (∠ ABD = ∠ ACD in the same segment)
= AC / sin∠ADC ..... ( By sine formula in △ADC)
= AC / sin 90° ..........( ∠ in semi circle)
= AC∴ BD = AC sin(a + b) ..... (1)
A────────B
| a \ b
|.......\
|...........\
|...............\
|...................\
|......................\
D────────C
Note that ∠ADC = ∠ABC = 90° (∠s in semi circle) ,
by Ptolemy's theorem , AC * BD = AD * BC + CD * AB
BD = AC (AD/AC * BC/AC + CD/AC * AB/AC)
BD = AC ( cos a * sin b + sin a * cos b ) ..... (2)(1) & (2) gives sin(a + b) = cos a sin b + sin a cos b
2012-11-19 20:31:13 補充：
The first line is :
In Δ ABD , by sine formula ,
2012-11-20 18:38:36 補充：
You can omit them if you know a/sinA = 2R.
Thankyou~
