第23页

信息发布者：
（1）证明：因为$OC = OB，$所以$\angle BCO = \angle B。$又因为$\angle B = \angle D，$所以$\angle BCO = \angle D。$
（2）解：因为$AB$是$\odot O$的直径，且$CD\perp AB$于点$E，$所以$CE = \frac{1}{2}CD = \frac{1}{2}\times4\sqrt{2}=2\sqrt{2}。$设$\odot O$的半径为$r，$则$OC = r，$$OE = OA - AE = r - 2。$在$Rt\triangle OCE$中，根据勾股定理$OC^{2}=CE^{2}+OE^{2}，$即$r^{2}=(2\sqrt{2})^{2}+(r - 2)^{2}，$展开得$r^{2}=8+r^{2}-4r + 4，$移项化简得$4r = 12，$解得$r = 3。$所以$\odot O$的半径为$3。$
解: （1）在$\triangle AEB$和$\triangle DEC$中，因为$\angle A = \angle D，$$AE = DE，$$\angle AEB = \angle DEC，$所以$\triangle AEB\cong\triangle DEC$（$ASA$），所以$BE = CE。$又因为$BC = CE，$所以$BE = CE = BC，$所以$\triangle EBC$是等边三角形，所以$\angle ACB = 60^{\circ}。$
（2）过点$B$作$BM\perp AC$于点$M。$因为$OF\perp AC，$所以$AF = CF。$因为$\triangle EBC$是等边三角形，所以$\angle GEF = \angle BCM = 60^{\circ}，$所以$\angle EGF = 30^{\circ}。$因为$EG = 2，$所以$EF = 1。$又因为$DE = AE = 3，$所以$CF = AF = AE + EF = 4，$所以$AC = 8，$$CE = 5，$所以$BC = 5。$因为$\angle BCM = 60^{\circ}，$所以$\angle MBC = 30^{\circ}，$所以$CM=\frac{1}{2}BC=\frac{5}{2}。$根据勾股定理可得$BM = \sqrt{BC^{2}-CM^{2}}=\sqrt{25 - \frac{25}{4}}=\frac{5\sqrt{3}}{2}，$$AM = AC - CM = 8-\frac{5}{2}=\frac{11}{2}。$所以$AB = \sqrt{AM^{2}+BM^{2}}=\sqrt{(\frac{11}{2})^{2}+(\frac{5\sqrt{3}}{2})^{2}}=\sqrt{\frac{121}{4}+\frac{75}{4}}=\sqrt{\frac{196}{4}} = 7。$
（1）证明：连接$OE，$因为$OE = OB，$所以$\angle ABE = \angle OEB。$因为$\odot O$与$AC$边相切于点$E，$所以$AC\perp OE，$即$\angle OEC = 90^{\circ}。$又因为$\angle ACB = 90^{\circ}，$所以$\angle ACB+\angle OEC = 180^{\circ}，$所以$BC// OE，$所以$\angle CBE = \angle OEB，$所以$\angle ABE = \angle CBE，$即$BE$平分$\angle ABC。$
（2）解：过点$O$作$OL\perp BC$于点$L，$则$\angle OLC = \angle OLB = 90^{\circ}，$$BL = FL。$因为$\angle ACB = \angle OLC = \angle OEC = 90^{\circ}，$所以四边形$OLCE$是矩形，所以$CL = OE = OB，$$OL = CE = 4。$所以$BL = FL = CL - CF = OB - 2。$在$Rt\triangle OLB$中，根据勾股定理$OL^{2}+BL^{2}=OB^{2}，$即$4^{2}+(OB - 2)^{2}=OB^{2}，$展开得$16+OB^{2}-4OB + 4 = OB^{2}，$移项化简得$4OB = 20，$解得$OB = 5。$所以$\odot O$的半径为$5。$
上一页 下一页