《2025年多维导学案九年级数学全一册人教版》

答案： 圆心
@@$AB = CD$ $\angle 1=\angle 2$$AB = CD$ $\overset{\frown}{AB}=\overset{\frown}{CD}$$\overset{\frown}{AB}=\overset{\frown}{CD}$ $\angle 1=\angle 2$

【例1】(人教教材母题)如图,$AB$是$\odot O$的直径,$\overset{\frown }{BC}=\overset{\frown }{CD}=\overset{\frown }{DE},∠COD=35^{\circ }$.求$∠AOE$的度数.

答案： 解：$\because \overset{\frown}{BC}=\overset{\frown}{CD}=\overset{\frown}{DE}$，$\angle COD = 35^{\circ}$，
$\therefore \angle BOC=\angle EOD=\angle COD = 35^{\circ}$．
$\therefore \angle AOE = 180^{\circ}-3\angle COD = 75^{\circ}$．

【变式1】如图,已知$AB,CD$是$\odot O$的直径,$\overset{\frown }{AE}=\overset{\frown }{AC},∠BOD=32^{\circ }$,求$∠COE$的度数.

答案： 解：$\because \overset{\frown}{AE}=\overset{\frown}{AC}$，
$\therefore \angle AOE=\angle AOC$．
$\because \angle AOC=\angle BOD = 32^{\circ}$，
$\therefore \angle AOE = 32^{\circ}$．
$\therefore \angle COE = 2\angle AOE = 64^{\circ}$．

【变式2】如图,$OA,OB,OC$是$\odot O$的半径,$\overset{\frown }{AC}=\overset{\frown }{BC},CD⊥OA$于点$D,CE⊥OB$于点$E$.求证:$AD=BE$.

答案： 证明：$\because \overset{\frown}{AC}=\overset{\frown}{BC}$，
$\therefore \angle AOC=\angle BOC$．
$\therefore OC$平分$\angle AOB$．
$\because CD\perp OA$，$CE\perp OB$，
$\therefore CD = CE$．
$\because OC = OC$，
$\therefore Rt\triangle COD\cong Rt\triangle COE$．
$\therefore OD = OE$．
$\because OA = OB$，
$\therefore OA - OD = OB - OE$．
$\therefore AD = BE$．