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F.4 circles

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1)http://img695.imageshack.us/img695/1861/dsc00577k.jpghttp://img97.imageshack.us/img97/3607/dsc00578qt.jpgans: b)1212)http://img718.imageshack.us/img718/7500/dsc00580r.jpghttp://img340.imageshack.us/img340/7234/dsc00581io.jpgans:ai)60... 顯示更多 1) http://img695.imageshack.us/img695/1861/dsc00577k.jpg http://img97.imageshack.us/img97/3607/dsc00578qt.jpg ans: b)121 2) http://img718.imageshack.us/img718/7500/dsc00580r.jpg http://img340.imageshack.us/img340/7234/dsc00581io.jpg ans: ai)60 bi)60 3) http://img29.imageshack.us/img29/3913/dsc00582jf.jpg http://img192.imageshack.us/img192/1406/dsc00583he.jpg ans:a)34 pls show clear steps for the proof and calculation.thz!

最佳解答:

1a) ∠ADG = ∠GEB (ext. ∠, cyclic quad.) ∠AFG = ∠GEC (ext. ∠, cyclic quad.) ∠GEB + ∠GEC = 180 So ∠ADG + ∠AFG = 180 and hence A, D, G and F are concyclic (Converse of opp. ∠s, cyclic quad.) b) ∠BAC = 180 - 64 - 57 = 59 Since ADGF is a cyclic quad., ∠BAC + ∠DGF = 180 and hence ∠DGF = 180 - 59 = 121 2a i) By tangent properties, ∠OCP = 90 and hence ∠POC = 180 - 30 - 90 = 60 ii) ∠COB = 120, also △OBC is an isos. △ with OC = OB, hence ∠OBC = 30 Also by tangent properties, ∠OPA = 30 and so ∠OPA = ∠CBP implies that CB//PAE b i), ii) ∠CPA = 60, also by tangent properties PC = PA and hence △CPA is an equil. △. Therefore ∠CAP = 60 3a) Since BE = EF, ∠FCE = ∠BCE = 34 (Equal chords, equal ∠s at circumference) b) ∠ABE = ∠ECB = 34 (∠ in alt. segment) ∠ABE = ∠FCE and hence ABCD is a cyclic quad. (Converse of ∠s in the same segment) c) ∠ADB = ∠ACB = 34 (∠ in the same segment) So ∠ADB = ∠ABD and hence △ABD is an isos. △

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