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$解：如圖，連 A D， B D$

$過(guò)點(diǎn) 作 B E \perp C D 于 E$

$\because A B 是直徑$

$\therefore \angle A C B=90^{\circ}，\angle A D B=90^{\circ}$

$\because A C=6， A B=10$

$\therefore B C=\sqrt{A B^2-A C^2}$

$=\sqrt{10^2-6^2}=8$

$\because C D 平分 \angle A C B$

$\therefore \angle B C D=45^{\circ}$

$\because B E \perp C D$

$\therefore C E=B E$

$\because C E^2+B E^2=B C^2， B C=8$

$\therefore C E=B E=4 \sqrt{2}$

$\because C D 平分 \angle A C B$

$\therefore\widehat{ A D}=\widehat{B D}$

$\therefore A D=B D$

$\because A D^2+B D^2=A B^2， A B=10$

$\therefore B D=5 \sqrt{2}$

$\therefore 在 R t \triangle B D E 中，$

$\because B D=5 \sqrt{2}， B E=4 \sqrt{2}$

$\therefore D E=\sqrt{B D^2-B E^2}$

$=\sqrt{(5 \sqrt{2})^2-(4 \sqrt{2})^2}=3 \sqrt{2}$

$\therefore C D=C E+D E$

$=4 \sqrt{2}+3 \sqrt{2}=7 \sqrt{2}$

$解：(1)∠ ABC與\angle C相等$

$連接OD$

$\because FD切\(zhòng)odot O于點(diǎn)D$

$\therefore OD\bot DF$

$\because BF\bot DF$

$\therefore OD//BF$

$\therefore \angle ODB=\angle FBD$

$\because OB=OD$

$\therefore \angle ODB=\angle OBD$

$\therefore \angle OBD=\angle FBD$

$\because AC//BF$

$\therefore \angle C=\angle FBD$

$\therefore \angle C=\angle OBD$

$∴\angle ABC=\angle C$

$(2)連接BE，OE$

$\because \widehat{DG}={60}^{\circ }$

$\therefore \angle GOD={60}^{\circ }$

$\therefore \angle FBD=\frac {1} {2}\angle GOD={30}^{\circ }$

$\because AC//BF$

$\therefore \angle C=\angle FBD={30}^{\circ } $

$由 ( {1} )結(jié)論\angle ABC=\angle C得$

$\angle ABC={30}^{\circ }$

$\because \angle BAE=\angle C+\angle ABC$

$\therefore \angle BAE={60}^{\circ }$

$\because OA=OE$

$\therefore \triangle OAE是等邊三角形$

$\therefore \angle EOA={60}^{\circ }$

$\therefore \angle ABE=\frac {1} {2}\angle EOA={30}^{\circ }$

$\therefore \angle ABE=\angle ABC$

$\therefore \widehat{AD}=\widehat{AE}$

$\therefore 點(diǎn)D與點(diǎn)E關(guān)于直線AB對(duì)稱(chēng)$

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第122頁(yè)