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1題 f.4 maths question 10M
發問:
圖片參考：http://img36.imageshack.us/img36/7471/64791566.jpg please teach me how to prove it i have tried many times but still cannot thank you picture website:http://img36.imageshack.us/img36/7471/64791566.jpg
最佳解答:
a) By the given △BDE ≡ △FDC , we have :BD = FD ......(1)CD = ED ......(2)ㄥBDE = ㄥFDC ㄥBDE - ㄥFDE = ㄥFDC - ㄥFDE ㄥBDF = ㄥCDE .......(3)By (1) , (2) , (3) : △BDF and △CDE both are isos. Δ with the same upper angle.So△BDF ~ △CDE (S.A.S.)Therefore ㄥFBD = ㄥCED since the base ㄥ of a pair similar isos. Δ are equal.And hence :ABDE is a cyclicquad (ext.∠= int. opp.∠) b)Similarly ,ㄥDCE = ㄥDFB ,CDFA is a cyclicquad (ext.∠= int. opp.∠)
1題 f.4 maths question 10M
發問:
圖片參考：http://img36.imageshack.us/img36/7471/64791566.jpg please teach me how to prove it i have tried many times but still cannot thank you picture website:http://img36.imageshack.us/img36/7471/64791566.jpg
最佳解答:
a) By the given △BDE ≡ △FDC , we have :BD = FD ......(1)CD = ED ......(2)ㄥBDE = ㄥFDC ㄥBDE - ㄥFDE = ㄥFDC - ㄥFDE ㄥBDF = ㄥCDE .......(3)By (1) , (2) , (3) : △BDF and △CDE both are isos. Δ with the same upper angle.So△BDF ~ △CDE (S.A.S.)Therefore ㄥFBD = ㄥCED since the base ㄥ of a pair similar isos. Δ are equal.And hence :ABDE is a cyclicquad (ext.∠= int. opp.∠) b)Similarly ,ㄥDCE = ㄥDFB ,CDFA is a cyclicquad (ext.∠= int. opp.∠)
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