如图所示,AB为圆O的直径,C为圆O上一点,AD和过C点的切线相交于点D,和圆O相交于点E,且AC平分∠DAB(1)求证:∠ADC=90°;(2)若AB=2r,AD=8/5r,求CD.
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![如图所示,AB为圆O的直径,C为圆O上一点,AD和过C点的切线相交于点D,和圆O相交于点E,且AC平分∠DAB(1)求证:∠ADC=90°;(2)若AB=2r,AD=8/5r,求CD.](/uploads/image/z/9285413-5-3.jpg?t=%E5%A6%82%E5%9B%BE%E6%89%80%E7%A4%BA%2CAB%E4%B8%BA%E5%9C%86O%E7%9A%84%E7%9B%B4%E5%BE%84%2CC%E4%B8%BA%E5%9C%86O%E4%B8%8A%E4%B8%80%E7%82%B9%2CAD%E5%92%8C%E8%BF%87C%E7%82%B9%E7%9A%84%E5%88%87%E7%BA%BF%E7%9B%B8%E4%BA%A4%E4%BA%8E%E7%82%B9D%2C%E5%92%8C%E5%9C%86O%E7%9B%B8%E4%BA%A4%E4%BA%8E%E7%82%B9E%2C%E4%B8%94AC%E5%B9%B3%E5%88%86%E2%88%A0DAB%281%29%E6%B1%82%E8%AF%81%EF%BC%9A%E2%88%A0ADC%3D90%C2%B0%EF%BC%9B%282%29%E8%8B%A5AB%3D2r%2CAD%3D8%2F5r%2C%E6%B1%82CD.)
如图所示,AB为圆O的直径,C为圆O上一点,AD和过C点的切线相交于点D,和圆O相交于点E,且AC平分∠DAB(1)求证:∠ADC=90°;(2)若AB=2r,AD=8/5r,求CD.
如图所示,AB为圆O的直径,C为圆O上一点,AD和过C点的切线相交于点D,和圆O相交于点E,且AC平分∠DAB
(1)求证:∠ADC=90°;
(2)若AB=2r,AD=8/5r,求CD.
如图所示,AB为圆O的直径,C为圆O上一点,AD和过C点的切线相交于点D,和圆O相交于点E,且AC平分∠DAB(1)求证:∠ADC=90°;(2)若AB=2r,AD=8/5r,求CD.
1、连接BC
∵AB是直径
∴∠ACB=90°
∵CD是圆切线
∴∠DCA=∠B
∵AC平分∠DAB
∴∠DAC=∠BAC
∴△ADC∽△ACB
∴∠ADC=∠ACB=90°
2、∵△ADC∽△ACB
∴AD/AC=AC/AB ,AD/AC=DC/BC
AC²=AB×AD=16/5r² ,AC=(4√5/5)r
∴BC²=AB²-AC²=4r²-16/5r²=4/5r²
BC=(2√5/5)r
∴CD=AD×BC/AC=(8/5)r×(2√5/5)r/(4√5/5)r=(4/5)r
(1) 连接OC
AC平分∠DAB
则 ∠DAC=∠CAB=∠ACO
CD为过C点的圆切线
OC⊥CD
∠ACO+∠ACD=90°
即 ∠DAC+∠ACD=90°
∴ ∠ADC=90°
(2) ∠DAC=∠CAB
∠ADC=∠ACB=90°
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(1) 连接OC
AC平分∠DAB
则 ∠DAC=∠CAB=∠ACO
CD为过C点的圆切线
OC⊥CD
∠ACO+∠ACD=90°
即 ∠DAC+∠ACD=90°
∴ ∠ADC=90°
(2) ∠DAC=∠CAB
∠ADC=∠ACB=90°
△ACD∽△ABC
AB/AC=AC/AD
AC^2=AB*AD=2r*(8/5)r=(16/5)r^2
CD^2=AC^2-CD^2=(16/25) r^2
CD=4r/5
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