第62页 - 苏科版九年级上下册数学书答案教科书答案课本答案

解：$\angle DAE$与$\angle DAC$相等。

理由如下：

$\because$四边形$ABCD$内接于$\odot O$，

$\therefore \angle DCB + \angle DAB = 180°$

又$\because \angle DAB + \angle DAE = 180°$

$\therefore \angle DCB = \angle DAE$

在$\triangle DBC$中，$\because DB = DC$，

$\therefore \angle DCB = \angle DBC.$

$\therefore \angle DBC = \angle DAE.$

又$\because \angle DBC = \angle DAC$，

$\therefore \angle DAE = \angle DAC.$

$解：EF垂直平分BC．理由如下：$

$∵AF平分∠BAD，AE平分∠BAC$

$∴∠BAF=\frac {1}{2}∠BAD，∠BAE=\frac {1}{2}∠BAC$

$∴∠BAF+∠BAE=\frac {1}{2}(∠BAD+∠BAC)=\frac {1}{2}×180°=90°，即∠EAF=90°$

$∴EF 为⊙O的直径$

$∵AE平分∠BAC$

$∴∠BAE=∠CAE$

∴ $\widehat{BE}=\widehat{CE}$

∴EF垂直平分BC

解：设$\angle A$，$\angle B$，$\angle C$的度数分别为$3x$，$4x$，$6x$。

由圆内接四边形的性质知$3x + 6x = 180°$，

解得$x = 20°$。

$\therefore \angle A = 3x = 60°$，$\angle B = 4x = 80°$，$\angle C = 6x = 120°$，

$\therefore \angle D = 180° - 80° = 100°$。

解：在$\odot O$的内接四边形$ABCD$中，

$\because \angle BAD = 60°$。

$\therefore \angle BCD = 180° - \angle BAD = 180° - 60° = 120°$

$\because \angle ACB = 70°$，

$\therefore \angle ACD = \angle BCD - \angle ACB = 120° - 70° = 50°$

$\therefore \angle ABD = \angle ACD = 50°$。

解：在$\odot O$的内接四边形$ABCD$中，

$\because \angle BAD = 60°$。

$\therefore \angle BCD = 180° - \angle BAD = 180° - 60° = 120°$

$\because \angle ACB = 70°$，

$\therefore \angle ACD = \angle BCD - \angle ACB = 120° - 70° = 50°$

$\therefore \angle ABD = \angle ACD = 50°$。

解：$AB$是半圆的直径，

$\therefore \angle ACB = 90°$

$\because \angle BAC = 20°$，

$\therefore \angle B = 90° - \angle BAC = 90° - 20° = 70°$

$\because$四边形$ABCD$内接于半圆，

$\therefore \angle D = 180° - \angle B = 180° - 70° = 110°$。

$\because \overgroup{AD} = \overgroup{CD}$，

$\therefore \angle DCA = \frac{1}{2}\angle B = \frac{1}{2} × 70° = 35°$。

$\therefore \angle DCB = \angle ACB + \angle DCA = 90° + 35° = 125°$，$\angle DAB = 180° - 125° = 55°$。