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楼主 |
发表于 2026-8-19 16:15
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Proof 1
Consider, for example, the case where E and F are located
outside the square ABCD. First of all, triangles CDE and
and CBF are equal.
Indeed, CD = CB and DE = BF. Also, ∠CDE = 90° + 60° = ∠CBF.
Next, ∠DCE + ∠BCF = ∠DCE + ∠DEC = 180° - 150° = 30°.
Therefore, in CEF, ∠ECF = 90° - 30° = 60°
and also CE = CF. CEF is thus equilateral. |
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