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maths mc a_a 要詳解
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(1a) Join OB and BC, then ∠OBA = 90° ∠BOC = 120° (Ext. angle of triangle) ∠OBC = ∠OCB = 30° (Base angles of isos. triangle) (1b) (i) ∠DAC = 30° (Tangent properties) ∠DAB + ∠DCB = 180° (Opp. angles of cyclic quad.) ∠DCB = 120° ∠DCA = 90° So, DC is a tangent to the circle at C. (ii) DE = DC = 5 (Tangent properties) AE = AB = 8 (Tangent properties) By Pyth. theorem, AC = 12 So, OA = 8. (2a) Join AB and BD, then ∠ACD = ∠BDC = ∠CAB = ∠DBA for the reasons of alternate angles AB//DC and angle in the same segment. Therefore AD = BC for the reason of equal angle at circumference means equal chord. (2b) Drop perpendiculars from P and Q to SR and let their feet of perpendicular be T and U respectively, then PT = QU (Intercept theorem) ∠PTS = ∠QUR = 90° So PTS and QTR are congurent triangles (RHS) Then ∠QRS + ∠SPT = 90° which gives ∠QRS + ∠SPQ = 180° So, P, Q, R and S are concyclic. (3a) Join BC, then ∠BCA = 90° (Angle in semi-circle) ∠BCP = ∠BAC (Angle in alt. segment) ∠APC = ∠BAC (Base angles of isos. triangle) ∠ABC = ∠BPC + ∠BCP (Ext. angle of triangle) 90° - ∠BAC = 2 ∠BAC ∠BAC = 30° (3b) ∠PCD = ∠APC + ∠PAC = 60° (Ext. angle of triangle) PC = PD (Tangent properties) ∠PDC = ∠PCD = 60° (Base angle of isos. triangle) So PDC is an isos. triangle and therefore, CD = PC = CA So ABC and CDE are equal circles. (4a) ∠AEF = ∠ADF = 90° ∠AEF + ∠ADF = 180° So, A, E, F and D are concyclic. (4b) ∠BEC = ∠BDC = 90° So, B, C, D and E are concyclic (converse of angle in the same segment). (4c) From the results of (a) and (b): ∠FAD = ∠FED = ∠CBD for the reason of angle in the same segment.. (4d) Extend AF to meet BC at G, then AFD and BCD are similar triangles (AAA) So, ∠AFD = ∠BCD Therefore AFD is also similar to ACG (AAA) Finally ∠AGC = ∠ADF = 90°, i.e. AF is perpendicular to BC.
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