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標題:
form 4 math circle
發問:
plz help me to solve the math questions below!thx so much!!!! Q 36 --> http://img59.imageshack.us/img59/4316/mathupload3.jpg Q 37-39 --> http://img59.imageshack.us/img59/2679/mathupload4.jpg
最佳解答:
(36a) With ∠TBO = ∠TAO = 90 (since TB and TA are tangents), we have: ∠TBO + ∠TAO = 180 Hence, T, B, O and A are concyclic. (b) With tangent properties, we can deduce that △TBO and △TAO are congurent. So let ∠BTO = ∠ATO = θ, then ∠BOD = 2θ (Ext∠ of cyclic quad.) Also, ∠ODB = (180 - 2θ)/2 = 90 - θ Then ∠TOA = 90 - θ = ∠ODB Thus, BD//TO (c) Since TB = TA, △TAB is an isos. △ with ∠BTA = 2θ Hence △TAB ~ △OBD (37a) With ∠s in alt. segment, we have: ∠TQM = x and ∠TQN = y Hence ∠MQN = x + y (b) Consider △TMN, ∠MTN = 180 - x - y, hence ∠MTN +∠MQN = 180 and therefore Q, M, T and N are concyclic. (c) Since Q, M, T and N are concyclic, ∠MNT = ∠TQM = x (∠s in the same segment) Hence ∠MNT = ∠MPR, giving P, M, N and R are concyclic. (Converse of ext. ∠ of cyclic quad.) (38a) ∠ATR = ∠PQT and ∠ATR = ∠ABT (∠s in alt. segment) Thus ∠PQT = ∠ABT and hence PQ//AB (b) With AB//SQ, ∠BSQ = ∠SQP (Alt. ∠s) Also ∠SQP = ∠STP (∠s in the same segment) So ∠STP = ∠BSQ Also ∠STQ = ∠BSQ (∠ in alt. segment) So ∠STQ = ∠STP and hence ST bisects ∠ATB. (c) ∠AST = ∠SQT (∠ in alt. segment) With ∠STQ = ∠ATS (proved in b part), we have △STQ ~ △ATS (AAA) (39a) ∠ADB = 90 (∠ in semi-circle) So ∠ADC = 90 Therefore △ABD and △ACD are congurent (RHS) (b) Since △ABD and △ACD are congurent, ∠DAB = ∠DAC = ∠DAE. Also, ∠EDA = ∠DBA (∠ in alt. segment) So △ABD ~ △ADE (AAA) (c) (i) By Pyth. thm, AD = 3 and hence since △ABD ~ △ADE: AB/AD = BD/DE 5/3 = 4/DE DE = 12/5 (ii) Join BF and then ∠BFA = 90 (Angle in semi-circle) So △CAD ~ △CBF (AAA) CA/CB = CD/CF 5/8 = 4/CF CF = 32/5 AF = CF - CA = 7/5
其他解答:
form 4 math circle
發問:
plz help me to solve the math questions below!thx so much!!!! Q 36 --> http://img59.imageshack.us/img59/4316/mathupload3.jpg Q 37-39 --> http://img59.imageshack.us/img59/2679/mathupload4.jpg
最佳解答:
(36a) With ∠TBO = ∠TAO = 90 (since TB and TA are tangents), we have: ∠TBO + ∠TAO = 180 Hence, T, B, O and A are concyclic. (b) With tangent properties, we can deduce that △TBO and △TAO are congurent. So let ∠BTO = ∠ATO = θ, then ∠BOD = 2θ (Ext∠ of cyclic quad.) Also, ∠ODB = (180 - 2θ)/2 = 90 - θ Then ∠TOA = 90 - θ = ∠ODB Thus, BD//TO (c) Since TB = TA, △TAB is an isos. △ with ∠BTA = 2θ Hence △TAB ~ △OBD (37a) With ∠s in alt. segment, we have: ∠TQM = x and ∠TQN = y Hence ∠MQN = x + y (b) Consider △TMN, ∠MTN = 180 - x - y, hence ∠MTN +∠MQN = 180 and therefore Q, M, T and N are concyclic. (c) Since Q, M, T and N are concyclic, ∠MNT = ∠TQM = x (∠s in the same segment) Hence ∠MNT = ∠MPR, giving P, M, N and R are concyclic. (Converse of ext. ∠ of cyclic quad.) (38a) ∠ATR = ∠PQT and ∠ATR = ∠ABT (∠s in alt. segment) Thus ∠PQT = ∠ABT and hence PQ//AB (b) With AB//SQ, ∠BSQ = ∠SQP (Alt. ∠s) Also ∠SQP = ∠STP (∠s in the same segment) So ∠STP = ∠BSQ Also ∠STQ = ∠BSQ (∠ in alt. segment) So ∠STQ = ∠STP and hence ST bisects ∠ATB. (c) ∠AST = ∠SQT (∠ in alt. segment) With ∠STQ = ∠ATS (proved in b part), we have △STQ ~ △ATS (AAA) (39a) ∠ADB = 90 (∠ in semi-circle) So ∠ADC = 90 Therefore △ABD and △ACD are congurent (RHS) (b) Since △ABD and △ACD are congurent, ∠DAB = ∠DAC = ∠DAE. Also, ∠EDA = ∠DBA (∠ in alt. segment) So △ABD ~ △ADE (AAA) (c) (i) By Pyth. thm, AD = 3 and hence since △ABD ~ △ADE: AB/AD = BD/DE 5/3 = 4/DE DE = 12/5 (ii) Join BF and then ∠BFA = 90 (Angle in semi-circle) So △CAD ~ △CBF (AAA) CA/CB = CD/CF 5/8 = 4/CF CF = 32/5 AF = CF - CA = 7/5
其他解答:
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