黑的发靓
幼苗
共回答了16个问题采纳率：93.8% 举报
连接CF,则CF⊥AE
∵BE⊥AE
∴CF∥BE
∴AF/AE = CF/BE = AC/AB
设OC = r,则AB = 4r
∵AE = 8
∴AF = 6,EF = 2
△ACF勾股定理得
AC² - CF² = AF²
即（3r）² - r² = 6²
∴r = 3√2/2,即CF = 3√2/2
∴BE = 2√2
S△BCE = S△ABE - S△ACE
= 1/2(AExBE - AExCF)
= 1/2AE(BE - CF)
= 1/2x8x(2√2 - 3√2/2)
= 2√2
1年前
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