△ABD中,正弦定理得AB/sinADB=BD/sin∠1同理,有AC/sinADC=AC/sin(180°-∠ADB)=AC/sinADB=CD/sin∠2两式相除,有AB/AC=BDsin∠2/CDsin∠1∵∠1=∠2,∴sin∠1=sin∠2
∴AB/AC=BD/CD设∠ADB=∠3正弦定理得AB/sin∠3=BD/sin(180°-∠2)=BD/sin∠2同理,AC/sin∠3=CD/sin∠1相除,得AB/AC=BDsin∠1/CDsin∠2∵∠1=∠2,∴sin∠1=sin∠2∴AB/AC=BD/CD