视频 BV1c34y1a7D3 定理2. 证法3.
过B作AC平行线
交AD延长线于E
延长AB于F
使BF=AB
设
AB=a，AC=b，BD=c，CD=d
有
ad=bc
即
b²-d²=b²-bcd/a
即
(b²-d²)a/b=ab-cd
即
(b²/d²-1)/(ab)·a²d²=ab-cd
即
((a+b）²/(c+d）²-1)/(ab)·a²d²=ab-cd
即
a²d²/(c+d)²·((a+b)²-(c+d)²）/(ab)=ab-cd
即
AD²
=
d²/(c+d)²
4a²·(1+(a²+b²-(c+d）²）/(2ab）)/2
=
ab-cd
即
AD²=AB·AC-BD·CD
得证
