Question

AD是Rt△ABC的角平分线,∠C＝90度,求证：AC²除以AD²＝BC除以2BD

Answer 1

设AC=b,AB=c,BC=a,

∵AD是Rt△ABC的角平分线,

∴b/c=CD/BD=CD/(a-CD)=(a-BD)/BD

∴CD=ab/(b+c),BD=ac/(b+c)

∴2BD/BC=2c/(b+c)

∵AD²=AC²+DC²,

∴AD²/AC²=(AC²+DC²)/AC²=1+DC²/AC²=1+[ab/(b+c)]²/b²=1+a²/(b+c)²=[(b+c)²+a²]/(b+c)²

=(a²+b²+c²+2bc)/(b+c)²=(2c²+2bc)/(b+c)²=2c/(b+c)

∴AC²/AD²＝BC/2BD