第47页 - 亮点给力提优课时作业本九年级数学江苏版

$40^{\circ}$

$2\sqrt{3}-2$

D

$3$

 （1）证明：因为四边形$ABCD$内接于$\odot O，$所以$\angle ABC + \angle ADC = 180^{\circ}。$

 因为$\angle ABC = 60^{\circ}，$所以$\angle ADC = 180^{\circ} - \angle ABC = 120^{\circ}。$

 因为$DB$平分$\angle ADC，$所以$\angle ADB = \angle CDB = \frac{1}{2}\angle ADC = 60^{\circ}，$所以$\angle ACB = \angle ADB = 60^{\circ}，$$\angle BAC = \angle CDB = 60^{\circ}，$所以$\angle ABC = \angle ACB = \angle BAC，$所以$\triangle ABC$是等边三角形。

 （2）解：过点$A$作$AE\perp CD，$交$CD$的延长线于点$E，$过点$B$作$BF\perp AC$于点$F，$则$\angle AED = \angle AFB = 90^{\circ}。$

 因为$\angle ADC = 120^{\circ}，$所以$\angle ADE = 180^{\circ} - \angle ADC = 60^{\circ}，$所以$\angle DAE = 90^{\circ} - \angle ADE = 30^{\circ}，$所以$DE = \frac{1}{2}AD。$

 因为$AD = 2，$所以$DE = 1，$所以$AE = \sqrt{AD^{2}-DE^{2}}=\sqrt{3}。$

 因为$CD = 4，$所以$S_{\triangle ACD} = \frac{1}{2}CD\cdot AE = 2\sqrt{3}，$$CE = CD + DE = 5，$所以$AC = \sqrt{AE^{2}+CE^{2}} = 2\sqrt{7}。$

 因为$\triangle ABC$是等边三角形，所以$AB = AC = 2\sqrt{7}，$$AF = \frac{1}{2}AC = \sqrt{7}，$所以$BF = \sqrt{AB^{2}-AF^{2}}=\sqrt{21}，$所以$S_{\triangle ABC} = \frac{1}{2}AC\cdot BF = 7\sqrt{3}，$所以$S_{四边形ABCD} = S_{\triangle ACD} + S_{\triangle ABC} = 9\sqrt{3}。$

 （1）解：连接$AD，$$BE。$因为$AB$为$\odot O$的直径，所以$\angle ADB = \angle AEB = 90^{\circ}，$所以$AD \perp BC，$$BE \perp AC。$

 因为$AB = AC = 6，$所以$BD = CD = 2，$所以$BC = BD + CD = 4。$

 因为$AE^{2}+BE^{2}=AB^{2}，$$CE^{2}+BE^{2}=BC^{2}，$所以$AB^{2}-AE^{2}=BC^{2}-CE^{2}。$

 设$CE = x，$则$AE = AC - CE = 6 - x，$所以$6^{2}-(6 - x)^{2}=4^{2}-x^{2}，$

 $36-(36 - 12x + x^{2}) = 16 - x^{2}，$

 $36 - 36 + 12x - x^{2}=16 - x^{2}，$

 $12x = 16，$

 解得$x = \frac{4}{3}。$故$CE$的长为$\frac{4}{3}。$

 （2）$\angle BAC = 2\angle CBE。$

 证明：连接$AD。$因为$AB$为$\odot O$的直径，所以$\angle ADB = 90^{\circ}，$所以$AD \perp BC。$

 因为$AB = AC，$所以$\angle BAC = 2\angle CAD。$

 因为$\angle CBE = \angle CAD，$所以$\angle BAC = 2\angle CBE。$

 （3）相同。

 证明：连接$AD。$因为$AB$为$\odot O$的直径，所以$\angle ADB = 90^{\circ}，$所以$AD \perp BC。$

 因为$AB = AC，$所以$\angle BAC = 2\angle CAD。$

 因为四边形$AEBD$内接于$\odot O，$所以$\angle EAD + \angle CBE = 180^{\circ}。$

 因为$\angle EAD + \angle CAD = 180^{\circ}，$所以$\angle CBE = \angle CAD，$所以$\angle BAC = 2\angle CBE。$