如图已知四边形ABCD、CDEF、EFHG都是正方形求证∠ACB+∠AFB+∠AHB=90°

学习时间:2026-04-03 13:50:43 阅读：6639

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2026-04-03 13:50:43

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连接EH则∠AEH=∠ACF,AE/AC=EH/CF=根号2所以△AEH∽△ACF∴∠EAH=∠CAF∵∠EAF=∠AHB∴∠CAF=∠AHB∴∠AHB+∠AFB=∠CAF+∠AFB=∠ACB=45∴∠CAF+∠AFB+∠ACB=90

如图已知四边形ABCD、CDEF、EFHG都是正方形 求证∠ACB+∠AFB+∠AHB=90°