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[几何] 到线段PB,PD张角相等

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hbghlyj posted 2023-3-25 18:18 |Read mode

$∠BCP=∠DCP,∠BAP=∠DAP,∠PBA+∠PDA=∠BPD+∠BCD,$
求证: $∠BPC=∠DPA$

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到线段PB,PD张角相等.ggb (22.68 KB, Downloads: 51)

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2025-6-8 12:22 GMT+8

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