circle problem

[匿名]

2011-06-04 11:42 pm

chord PA and chord PB cut chord DC at points Q and R respectively and arc DP=arc PC. Let angle PBD=a and DCA=b

a i)Express DQA in terms of a and b
ii)Prove that ABRQ is a cyclic quadrilateral
b) If DBC=70 ,PRQ=80,find BDC

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回答 (1)

用戶25975

2011-06-05 1:08 am

✔ 最佳答案

a i) ∠QAC = ∠PAC = a (Equal arc, equal ∠ at circum)

∠DQA = a + b (Ext. ∠ of triangle)

ii) Join AB, we have ∠DBA = b (∠s in the same segment)

So ∠RBA = a + b = ∠DQA

So ABRQ is a cyclic quad. (Converse of ext. ∠ of cyclic quad.)

b) With ∠PBC = a, we hae ∠DBC = 2a and hence a = 35

So ∠DBR = 35 and so ∠BDC = ∠PRQ - a = 45

2011-06-04 17:54:56 補充：
Consider triangle DBR

∠BDC + ∠DBR = ∠PRQ (Ext. ∠ of triangle)

∠BDC + a = ∠PRQ

∠BDC = ∠PRQ - a

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