在△ABC中,D点是 ∠C的平分线与底边AB的交点,证明下列不等式:CD²=AC·BC-AD·BD.
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在△ABC中,D点是 ∠C的平分线与底边AB的交点,证明下列不等式:CD²=AC·BC-AD·BD.
在△ABC中,D点是 ∠C的平分线与底边AB的交点,证明下列等式:CD²=AC·BC-AD·BD.
可以用正弦定理证明:
设
x = 1/2 ∠C
y =∠ ADC
△ADC中
CD/sinA = AD/sinx = AC/siny = u
CD = u sinA = u sin(x+y)
AD = u sinx
AC = u siny
△BDC中
CD/sinB = BC/sin(π-y) = BD/sinx = v
CD = v sinB = v sin(x+(π-y)) = v sin(y-x)
BC = v siny
BD = v sinx
CD² = u sin(x+y) * v sin(y-x) = 1/2 uv [cos((x+y)+(y-x)) -cos((x+y)-(y-x))]
= 1/2 uv [ cos2y - cos2x]
= 1/2 uv [ (1-2sin²y) - (1-2sin²x) ]
= uv ( sin²y - sin²x )
AC*BC = u siny * v siny = uv sin²y
AD*BD = u sinx * v sinx = uv sin²x
所以
CD²=AC·BC-AD·BD
可以用正弦定理证明:
设
x = 1/2 ∠C
y =∠ ADC
△ADC中
CD/sinA = AD/sinx = AC/siny = u
CD = u sinA = u sin(x+y)
AD = u sinx
AC = u siny
△BDC中
CD/sinB = BC/sin(π-y) = BD/sinx = v
CD = v sinB = v sin(x+(π-y)) = v sin(y-x)
BC = v siny
BD = v sinx
CD² = u sin(x+y) * v sin(y-x) = 1/2 uv [cos((x+y)+(y-x)) -cos((x+y)-(y-x))]
= 1/2 uv [ cos2y - cos2x]
= 1/2 uv [ (1-2sin²y) - (1-2sin²x) ]
= uv ( sin²y - sin²x )
AC*BC = u siny * v siny = uv sin²y
AD*BD = u sinx * v sinx = uv sin²x
所以
CD²=AC·BC-AD·BD
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