作射線BO和外接圓交於K
連接KA,KC
KC和AH都垂直BC,所以KC//AH
KA和CH都垂直AB,所以KA//CH
故KAHC為平行四邊形
KC=AH=AD
KO=AO=AE
角OKC=角EAD
故三角形OKC和三角形EAD全等
DE=OC=OA=AE
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由 亞斯 於 星期一 十二月 04, 2006 10:40 am
[問題]幾何題(48)
由 tangpakchiu 於 星期五 十二月 01, 2006 6:22 am
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