Got this proof from the internet. www.ddhw.com As before, ABC is the triangle, BY, CZ the bisectors of B, C, with Y on AC, Z on AB. We have BY = CZ. Let E be the intersection of BY and CZ. Connect A, E. Find the point D outside of edge AB so that BD = AZ and DY = AC. www.ddhw.com So triangle DBY is congruent with AZC. Let the bisector of A, D, B, Y are concyclic ( www.ddhw.com Then A, D, F, E are concyclic ( Since AE = DF (DBY congruent with AZC) , we have AD // EF. Therefore,
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哈哈哈哈 O(∩_∩)O哈哈~ 好久不来了，不小心看了一眼自己主页，发现自动关联了自己以往发过的帖，点进这个看了一眼，笑个半死。。。敢情我当初还发过这么搞笑的东西啊，我自己完全不记得了 
雷达卡
发表于 2004-12-17 01:15:29
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