再提供三種作法
第一種
連CO_1
因為AC是直徑
所以角AO_1C=CO_1D=90度
又AO_1=D_1
所以三角形CO_1A全等於三角形CO_1D(SAS)
所以AC=CD
第二種
延長O_1O交圓O於K
由圓的性質可以知道
弧BO_1=弧AO_1=弧CK
角ADC=(弧AC-弧BO_1)/2
=(180度-弧AO_1)/2
=(弧KO_1-弧CK)/2
=弧CO_1/2=角CAD
所以CA=CD
第三種
延長O_1O交圓O於K
由圓的性質可以知道
弧BO_1=弧AO_1=弧CK
角KOC=角COB
所以KO_1平行CD
角ADC=角AO_1K=角CAD
所以CA=CD
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由 ☆ ~ 幻 星 ~ ☆ 於 星期日 七月 30, 2006 3:12 pm
由 宇智波鼬 於 星期五 七月 28, 2006 2:19 pm
[問題]幾何題32
由 tangpakchiu 於 星期五 七月 28, 2006 1:41 pm
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