數學~證明題目~

ㄥBCD + ㄥBFD = 90 + 90 = 180°
BCDF is a cyclic quadrilaterals. (opp. ∠s supp.)
Join CF and BD.
ㄥDBF = ㄥDCF (∠s in the same segment) .....(1)DF = DF (Common)
AF = BF (Given)
ㄥDFA = ㄥDFB = 90° (Given)△DFA ~= △DFB (S.A.S.)
So
ㄥDBF = ㄥDAF ......(2)ㄥEBF = ㄥABE (Common)
ㄥEFB = ㄥACB = 90° (Given)△EBF ~ △ABC (A.A.)
So
ㄥDAF = ㄥFEC .......(3)(1) & (2) :
ㄥDCF = ㄥDAF ......(4)(3) & (4) :ㄥDCF = ㄥFEC ,ㄥDFC = ㄥCFE (Common)△FCD ~ △FEC (A.A.)


2011-01-28 19:40:18 補充：
如果你未學四點共圓請參考以下證明 :

ㄥADF = ㄥEDC (對頂角)
ㄥAFD = ㄥECD = 90° (已知)
△ADF ~ △EDC (A.A.)
故
ㄥDAF = ㄥDEC (相似△對應角相等) 及
DA : DE = DF : DC (相似△對應邊成比例)

DA : DE = DF : DC (已證)
ㄥADE = ㄥFDC (對頂角)
故
△ADE ~ △FDC (S.A.S.)

得ㄥAED = ㄥDCF (相似△對應角相等) ........(1)

2011-01-28 19:40:48 補充：
EF = EF (公共)
ㄥEFA = ㄥEFB = 90° (已知)
AF = BF (已知)
故
△EFA ~= △EFB (S.A.S.)

得ㄥAED = ㄥCEF (全等△對應角相等) .......(2)

綜合 (1) , (2) :
ㄥDCF = ㄥCEF
又 ㄥDFC = ㄥCFE (公共)

故△FCD ~ △FEC (A.A.)