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[几何] 看见个简单的两圆证角等

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isee Posted 2021-9-30 00:33 |Read mode
Last edited by isee 2021-9-30 00:39如图,当然,竞赛中最简单的
c30.jpg

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乌贼 Posted 2021-9-30 01:12
1111-1.png

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 Author| isee Posted 2021-9-30 08:46
Last edited by isee 2021-9-30 09:35回复 2# 乌贼


    应该不垂直也成立吧,AX应该不会巧合为高——高与直径形成这两角是平凡的,图中是一种推广

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乌贼 Posted 2021-9-30 13:14
回复 3# isee
用几何画板验证一下,还真是 1111-2.png

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kuing Posted 2021-9-30 15:06
回复 4# 乌贼

嗯,确实如此:
QQ截图20210930152531.png
`D`, `E` 为 `BC` 上两点,`\triangle ABC` 的外接圆 `O_1` 与 `\triangle ADE` 的外接圆 `O_2` 交于另一点 `F`,`FD` 交圆 `O_1` 于 `G`,求证:`\angle EAB=\angle GAC`。

证明:`\angle ABE=\angle AGC` 且 `\angle AEB=180\du-\angle AFD=\angle ACG`,所以 `\triangle ABE\sim\triangle AGC`。

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 Author| isee Posted 2021-9-30 15:49
然后极限一下,主楼就成立了,哈哈哈哈哈

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