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第31页

信息发布者：

 解：

 (1) 作法：

 ① 作直径$AC；$

 ② 作直径$BD ⊥ AC；$

 ③ 依次连接$A,B,C,D$四个点，四边形$ABCD$即为$\odot O$的内接正方形；

 ④ 分别以点$A,C$为圆心，$OA$长为半径作弧，交$\odot O$于点$E,H,F,G；$

 ⑤ 顺次连接$A,E,F,C,G,H$各点，六边形$AEFCGH$即为$\odot O$的内接正六边形。

 (2) 证明：连接$OE,DE。$

 $\because ∠ AOD = \frac{360°}{4}=90°，$$∠ AOE = \frac{360°}{6}=60°，$

 $\therefore ∠ DOE = ∠ AOD - ∠ AOE = 90° - 60° = 30°。$

 $\therefore DE$为$\odot O$的内接正十二边形的一边。

 证明：

 (1) $\because \odot O$是$△ ABC$的内切圆，

 $\therefore BO$平分$∠ ABC，$$CO$平分$∠ ACB。$

 $\because DE // BC，$

 $\therefore ∠ DOB = ∠ OBC，$$∠ EOC = ∠ OCB。$

 $\therefore ∠ DBO = ∠ DOB，$$∠ ECO = ∠ EOC。$

 $\therefore BD = DO，$$EO = EC。$

 $\therefore BD + CE = DO + OE = DE。$

 (2) 解：$\because ∠ ABC,∠ ACB$的平分线交于点$O，$

 $\therefore ∠ OBC = \frac{1}{2}∠ ABC，$$∠ OCB = \frac{1}{2}∠ ACB。$

 $\because ∠ ABC + ∠ ACB = 180° - ∠ BAC = 110°，$

 $\therefore$ 在$△ OBC$中，

 $∠ BOC = 180° - ∠ OBC - ∠ OCB = 180° - \frac{1}{2}(∠ ABC + ∠ ACB) = 125°。$

 解：

 (1) $DE$与$\odot O$相切，理由如下：

 连接$OE。$

 $\because OB = OE，$

 $\therefore ∠ OBE = ∠ OEB。$

 $\because CD$是$△ ABC$的高，

 $\therefore ∠ CDB = 90°。$

 $\therefore ∠ DCB + ∠ DBC = 90°。$

 $\because ∠ DBC = ∠ OBE，$

 $\therefore ∠ DBC = ∠ OEB。$

 $\therefore ∠ DCB + ∠ OEB = 90°。$

 $\because DE = DC，$

 $\therefore ∠ DCB = ∠ DEB。$

 $\therefore ∠ DEO = ∠ DEB + ∠ OEB = 90°，$即$OE ⊥ DE。$

 $\because OE$为$\odot O$的半径，

 $\therefore DE$与$\odot O$相切。

 (2) 由

 (1)知，$∠ DEO = 90°。$

 设$BO = EO = AO = r，$

 $\because BD = 2，$

 $\therefore DO = 2 + r。$

 在$\mathrm{Rt}△ OED$中，$DE = DC = 4，$

 $DE^2 + EO^2 = DO^2，$即$4^2 + r^2 = (2 + r)^2，$

 解得$r = 3。$

 $\therefore DO = 5，$$AB = 6。$

 $\therefore AD = 8。$

 在$\mathrm{Rt}△ ACD$中，由勾股定理，得$AC = \sqrt{AD^2 + CD^2} = 4\sqrt{5}。$

 在$\mathrm{Rt}△ CDB$中，由勾股定理，得$BC = \sqrt{CD^2 + BD^2} = 2\sqrt{5}。$

 $\therefore △ ABC$的周长为$BC + AB + AC = 2\sqrt{5} + 6 + 4\sqrt{5} = 6 + 6\sqrt{5}。$

 解：

 (1) $\because ∠ BAE = ∠ CAD，$

 $\therefore ∠ BAE + ∠ BAD = ∠ CAD + ∠ BAD，$即$∠ EAD = ∠ BAC。$

 又$\because AD = AC，$$∠ ADE = ∠ ACB，$

 $\therefore △ ADE ≌ △ ACB。$

 $\therefore AE = AB。$

 $\because AB = 8，$

 $\therefore AE = 8。$

 (2) 证明：连接$BO$并延长，交$\odot O$于点$F，$连接$AF。$

 $\because BF$是$\odot O$的直径，

 $\therefore ∠ BAF = 90°。$

 $\therefore ∠ AFB + ∠ ABF = 90°。$

 $\because ∠ AFB = ∠ ACB，$

 $\therefore ∠ ACB + ∠ ABF = 90°。$

 在$△ ADC$中，$AD = AC，$

 $\therefore ∠ ACB = ∠ ADC。$

 $\therefore 2∠ ACB + ∠ CAD = 180°。$

由

 (1)知，$AE = AB，$

 $\therefore ∠ AEB = ∠ ABE。$

 $\therefore 2∠ ABE + ∠ BAE = 180°。$

 $\because ∠ BAE = ∠ CAD，$

 $\therefore ∠ ACB = ∠ ABE。$

 $\therefore ∠ ABE + ∠ ABF = 90°，$即$∠ OBE = 90°。$

 $\because OB$为$\odot O$的半径，

 $\therefore EB$是$\odot O$的切线。