Re: [中學] 一題幾何
請求高手幫忙，謝謝
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08/17 11:25,
08/17 11:25
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08/17 11:48,
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08/17 11:48,
08/17 11:48
△ABD ~ △CBF (AA~)
∠BAD = ∠1 = ∠BCF
A, E, D, B四點共圓 => ∠BED = ∠BAD = ∠1
B, F, E, C四點共圓 => ∠BEF = ∠BCF = ∠1
=> BE平分∠FED
對AC做△BAC鏡射△B'AC互為全等
D'為D之鏡射點，F'為F之鏡射點
BE⊥AC, B'E⊥AC => B, E, B'共線
因為∠F'ED' = ∠FED
B'E平分∠F'ED'
BE平分∠FED
=> ∠FEB = ∠B'ED'
=> F, E, D'共線
同理對AB做△CBA的鏡射全等三角形△C"BA
D"為D之鏡射點
仿上面證明可知D', F, D"三點共線
因此DE + EF + FD = D"F + FE + ED'
= 等腰三角形△AD"D'的底邊
又AD" = AD = AD'
∠D"AD' = ∠D"AB + ∠DAB + ∠CAD + ∠CAD'
= 2[∠DAB + ∠CAD] = 2∠A
=> DE + EF + FD = D"F + FE + ED'
= AD * sin(2∠A/2) * 2
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08/17 12:49, , 1F
08/17 12:49, 1F
※ 編輯: Honor1984 (61.56.10.112), 08/17/2017 14:32:33
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08/17 16:41, , 2F
08/17 16:41, 2F
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