中四圓形學問題 急急急

26)a)
ㄥBDC = 90°(∠in semi-circle)ㄥBCD = 180 - 32 - 90 = 58°b)
Join AC , then ㄥACB = ㄥACD (equal chords, equal ∠s)So ㄥACB = ㄥACD = 58/2 = 29°ㄥABD = ㄥACB = ㄥACD(equal chords, equal ∠s)= 29°
27)Join PQ and BD ,ㄥPQC = ㄥPBD (ext. ∠s, cyclic quad.)
ㄥPQS = ㄥPAS (∠s in the same segment)
ㄥAPR = 70° = ㄥBDR (ext. ∠s, cyclic quad.) In △ABD ,
ㄥPBD + ㄥPAS + ㄥBDR = 180° (∠ sum of △)
ㄥPQC + ㄥPQS + ㄥAPR = 180°
ㄥPQC + ㄥPQS + 70° = 180°
ㄥPQC + ㄥPQS = 110°
ㄥSQC = 110°
28)a)ㄥBAQ = ㄥPRB (∠s in the same segment)
ㄥPBQ = ㄥPBQ (common)△ABQ ~ △RBP (A.A.);ㄥCAR = ㄥSQC (∠s in the same segment)
ㄥSCR = ㄥSCR (common)△ACR ~ △QCS (A.A.)b)Since △ABQ ~ △RBP , we have :
AB : BR = BQ : BP
==>
AB x BP = BQ x BR ..........(1)And △ACR ~ △QCS , we have :
CA : CQ = CR : CS
==>
CA x CS = CR x CQ ..........(2)Comparing (1) and (2) ,
BQ = CR (given) , so
BR = BQ + QR = CR + QR = CQHence BQ x BR = CR x CQ ,
i.e. AB x BP = CA x CSc)ㄥPRQ = ㄥPSQ (∠s in the same segment) ㄥPSQ = ㄥSQR (alt.∠s, PS//BC)So ㄥPRQ = ㄥSQR.....(1) , arc PQ = arc SR (equal ∠s, equal arcs)Therefore
arc PQ + arc QR = arc SR + arc QR
arc PR = arc SQ
So
PR = SQ (equal arcs, equal chords)......(2)BR = CQ (proved in part b) .......(3)By (1) (2) (3) ,△BPR ~= △CQS (S.A.S.) ,BP = CSi.e. BP : CS = 1 : 1

2011-01-16 11:12:55 補充：
Alternative for Q27) :

Join PQ ,

ㄥSQC
= ㄥRDQ + ㄥQSD (ext. ∠ of Δ)
= ㄥRPQ (∠s in the same segment) + (180° - ㄥASQ)
= ㄥRPQ + (180° - ㄥAPQ(∠s in the same segment) )
= ㄥRPQ + (180° - (70°+ㄥRPQ))
= 110°

其餘那兩題呢 真的別無他法？