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春芽
共回答了23个问题采纳率:82.6% 举报
∵在Rt△ABC中,∠ACB=90°
∴∠CAB + ∠B =90°
∵CD⊥AB
∴∠CDF = ∠CDA = 90°
∴在Rt△CDB中,∠DCB + ∠B = 90°
∴ ∠CAB = ∠DCB
∵ AF平分∠CAB ,CG平分∠BCD
∴ ∠FAB= 1/2 ∠CAB ∠GCF = ∠DCG = 1/2 ∠DCB
即:∠FAB = ∠DCG
在Rt△ADE中,∠EAD + ∠AED = 90° ( ∠FAB = ∠EAD)
又∵ ∠AED = ∠CEF (对顶角)
∴ ∠DCG + ∠ CEF = 90°
∴ EF ⊥ CG
∴ ∠GCF + ∠ AFC= 90°
∴ ∠CEF = ∠AFC
∴ CF = CE
CG垂直平分EF
∠DCG + ∠EGC = ∠ DEG = ∠DCB
∠DCG = 1/2 ∠DCB
∠DCG = 1/2 ∠DCB
∠DCG = ∠EGC = 1/2 ∠DCB
EC = EG EF ⊥ CG
EF垂直平分CG
∴ 四边形CEGF是菱形 (对角线互相垂直平分
1年前
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