已知:如图,△ABC内接于⊙O,AB为⊙O的直径,∠CBA的平分线交AC于点F,交⊙O于点D,DE⊥AB于E,且交AC于如图,△ABC内接于O,AB为直径,∠CBA的平分线BD交AC于点已知:如图,△ABC内接于⊙O,AB为⊙O的直径,∠C
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![已知:如图,△ABC内接于⊙O,AB为⊙O的直径,∠CBA的平分线交AC于点F,交⊙O于点D,DE⊥AB于E,且交AC于如图,△ABC内接于O,AB为直径,∠CBA的平分线BD交AC于点已知:如图,△ABC内接于⊙O,AB为⊙O的直径,∠C](/uploads/image/z/8548428-12-8.jpg?t=%E5%B7%B2%E7%9F%A5%EF%BC%9A%E5%A6%82%E5%9B%BE%2C%E2%96%B3ABC%E5%86%85%E6%8E%A5%E4%BA%8E%E2%8A%99O%2CAB%E4%B8%BA%E2%8A%99O%E7%9A%84%E7%9B%B4%E5%BE%84%2C%E2%88%A0CBA%E7%9A%84%E5%B9%B3%E5%88%86%E7%BA%BF%E4%BA%A4AC%E4%BA%8E%E7%82%B9F%2C%E4%BA%A4%E2%8A%99O%E4%BA%8E%E7%82%B9D%2CDE%E2%8A%A5AB%E4%BA%8EE%2C%E4%B8%94%E4%BA%A4AC%E4%BA%8E%E5%A6%82%E5%9B%BE%2C%E2%96%B3ABC%E5%86%85%E6%8E%A5%E4%BA%8EO%2CAB%E4%B8%BA%E7%9B%B4%E5%BE%84%2C%E2%88%A0CBA%E7%9A%84%E5%B9%B3%E5%88%86%E7%BA%BFBD%E4%BA%A4AC%E4%BA%8E%E7%82%B9%E5%B7%B2%E7%9F%A5%EF%BC%9A%E5%A6%82%E5%9B%BE%2C%E2%96%B3ABC%E5%86%85%E6%8E%A5%E4%BA%8E%E2%8A%99O%2CAB%E4%B8%BA%E2%8A%99O%E7%9A%84%E7%9B%B4%E5%BE%84%2C%E2%88%A0C)
已知:如图,△ABC内接于⊙O,AB为⊙O的直径,∠CBA的平分线交AC于点F,交⊙O于点D,DE⊥AB于E,且交AC于如图,△ABC内接于O,AB为直径,∠CBA的平分线BD交AC于点已知:如图,△ABC内接于⊙O,AB为⊙O的直径,∠C
已知:如图,△ABC内接于⊙O,AB为⊙O的直径,∠CBA的平分线交AC于点F,交⊙O于点D,DE⊥AB于E,且交AC于
如图,△ABC内接于O,AB为直径,∠CBA的平分线BD交AC于点已知:如图,△ABC内接于⊙O,AB为⊙O的直径,∠CBA的平分线交AC于点F,交⊙O于点D,DE⊥AB于点E,且交AC于点P,连接AD.
(1)求证:PA=PD;
(2)求证:P是线段AF的中点.
(3)若圆O的半径为5,AF=15/2,求AD的长
已知:如图,△ABC内接于⊙O,AB为⊙O的直径,∠CBA的平分线交AC于点F,交⊙O于点D,DE⊥AB于E,且交AC于如图,△ABC内接于O,AB为直径,∠CBA的平分线BD交AC于点已知:如图,△ABC内接于⊙O,AB为⊙O的直径,∠C
(1)证明:∵BD平分∠CBA,
∴∠CBD=∠DBA,
∵∠DAC与∠CBD都是弧CD所对的圆周角,
∴∠DAC=∠CBD,
∴∠DAC=∠DBA,
∵AB是⊙O的直径,DE⊥AB,
∴∠ADB=∠AED=90°,
∴∠ADE+∠DAE=90°,∠DBA+∠DAE=90°,
∴∠ADE=∠DBA,
∴∠DAC=∠ADE,
∴PA=PD;
(2)
∵AB为直径,
∴∠ADB=90°,
∵DE⊥AB于E,
∴∠DEB=90°,
∴∠ADE+∠EDB=∠ABD+∠EDB=90°,
∴∠ADE=∠ABD=∠DAP,
∴PD=PA,
∵∠DFA+∠DAC=∠ADE+∠PDF=90°,且∠ADB=90°,
∴∠PDF=∠PFD,
∴PD=PF,
∴PA=PF,即P是线段AF的中点,
(3)
∵∠DAF=∠DBA,∠ADB=∠FDA=90°,
∴△FDA∽△ADB,
∴AD/DB =AF/BA ,
又圆O的半径为5,
∴AB=10,
∴AD/DB=15/2/10=3/4
设AD=3X,DB=4X
∵∠ADB=90°
∴AD^2+BD^2=AB^2
9x^2+16x^2=100
x^2=4
x=2
∴AD=6
图呢,上图!!!!问个问题都不专业,还想干啥?