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由 **☆ ~ 幻星 ~ ☆** 於星期日七月 23, 2006 2:05 pm

AB=CD=10
連HD
所以角CHD=90度
因為角COK=90度=角CHD，角KOC=角KOD
所以三角形CHD相似於三角形COK
KO：OC=DH：HC=3：4
OC=5
KO=5*3/4=15/4
BK：AK=(5+15/4)：(5-15/4)=7：1

由 **tangpakchiu** 於星期日七月 23, 2006 11:46 am

#ed_op#DIV#ed_cl##ed_op#IMG alt="image file name: 2kd765fb351c.gif" src="http://yll.loxa.edu.tw/phpBB2/richedit/upload/2kd765fb351c.gif" border=0#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#SPAN style="FONT-SIZE: 12pt; FONT-FAMILY: SimSun; mso-ascii-font-family: 'Times New Roman'; mso-hansi-font-family: 'Times New Roman'; mso-bidi-font-family: 新細明體; mso-ansi-language: EN-US; mso-fareast-language: ZH-TW; mso-bidi-language: AR-SA"#ed_cl##ed_op#/SPAN#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#SPAN style="FONT-SIZE: 12pt; FONT-FAMILY: SimSun; mso-ascii-font-family: 'Times New Roman'; mso-hansi-font-family: 'Times New Roman'; mso-bidi-font-family: 新細明體; mso-ansi-language: EN-US; mso-fareast-language: ZH-TW; mso-bidi-language: AR-SA"#ed_cl#在圓O中直徑AB，CD互相垂直，弦CH交AB於K，且AB=10cm，CH=8cm，求BK:AK的值。#ed_op#/SPAN#ed_cl##ed_op#/DIV#ed_cl#

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[quote="tangpakchiu"]#ed_op#DIV#ed_cl##ed_op#IMG alt="image file name: 2kd765fb351c.gif" src="http://yll.loxa.edu.tw/phpBB2/richedit/upload/2kd765fb351c.gif" border=0#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#SPAN style="FONT-SIZE: 12pt; FONT-FAMILY: SimSun; mso-ascii-font-family: 'Times New Roman'; mso-hansi-font-family: 'Times New Roman'; mso-bidi-font-family: 新細明體; mso-ansi-language: EN-US; mso-fareast-language: ZH-TW; mso-bidi-language: AR-SA"#ed_cl##ed_op#/SPAN#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#SPAN style="FONT-SIZE: 12pt; FONT-FAMILY: SimSun; mso-ascii-font-family: 'Times New Roman'; mso-hansi-font-family: 'Times New Roman'; mso-bidi-font-family: 新細明體; mso-ansi-language: EN-US; mso-fareast-language: ZH-TW; mso-bidi-language: AR-SA"#ed_cl#在圓O中直徑AB，CD互相垂直，弦CH交AB於K，且AB=10cm，CH=8cm，求BK:AK的值。#ed_op#/SPAN#ed_cl##ed_op#/DIV#ed_cl# [/quote]