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[几何] 等轴双曲线内接直角三角形 切线

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hbghlyj Posted 2021-6-1 04:18 |Read mode
等轴双曲线上三点A,B,C,AB⊥AC,求证:A处的切线⊥BC.

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isee Posted 2021-6-1 19:42
旋转成y=1/x,并不难计算呀

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isee Posted 2021-6-1 19:44
FAQ:等轴双曲线内接三角形的垂心在双曲线上

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isee Posted 2021-6-1 19:48

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 Author| hbghlyj Posted 2021-6-1 20:21
Last edited by 2023-3-25 18:37命题9的证明中的点S是什么呢
思路二:利用上题结论

解:如图,由上题知∠MAH=∠MZH,∠MHZ=180°-∠MDZ=180°-∠MSH=∠MSA,

∴∠AMD=∠ZMH,同理可得其余,故∠MHN,∠MDN相等或互补。
我的证明是
因为DAM-HMA=ADH+DHM=ZHD+DHM=ZHM=MDZ和MZD=DAM所以MZD-HMA=MDZ又MZD-MDZ=ZMD,所以ZMD=HMA

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 Author| hbghlyj Posted 2021-6-1 22:33
回复 7# ellipse

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