About circle 5 @ wszspj2的部落格

標題:

About circle 5

發問:

5 marks. please show steps. please answer. thank you. link: http://www.sendspace.com/file/uygv1l

最佳解答:

19b) Join AD and DC, then ∠TDA = ∠DCA (∠ in alt. segment) Also since CFDE is a cyclic quad., ∠DFE = ∠DCE (∠s in the same segment) Hence ∠DFE = ∠TDA So AD//EF 22b) ∠PQA = ∠QBA and ∠PRA = ∠RBA (∠s in alt. segment) Also ∠PQA = ∠RBA and ∠PRA = ∠QBA (∠s in the same segment) Thus ∠PQA = ∠PRA and hence △PQR is an isos. △ So PQ = PR 2010-07-27 23:58:24 補充： Let me think for a while more for Q23. Thanks. 2010-07-28 22:36:23 補充： 23a) By tangent prop., we have DC = DB. Thus ∠DCB = ∠DBC and we let them be θ. So ∠BAC = θ (∠ in alt. segment) Now, we let ∠DCE = ∠CFE = Φ (∠ in alt. segment) Then ∠CDE = 2θ (ext. ∠ of triangle) So, ∠CED = 180 - 2θ - Φ 2010-07-28 22:37:42 補充： Also, ∠CED = ∠CFE + ∠ECF (ext. ∠ of triangle) 180 - 2θ - Φ = Φ + ∠ECF ∠ECF = 180 - 2θ - 2Φ ∠ACB = 180 - ∠ECF - ∠ECB = 180 - (180 - 2θ - 2Φ) - (θ + Φ) = θ + Φ = ∠ECB Hence, △ABC ~ △BEC (AAA) 2010-07-28 22:38:29 補充： b) With △ABC ~ △BEC: BC/EC = AC/BC BC^2 = 16 BC = 4 cm