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F5 Maths
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圖片參考:http://imgcld.yimg.com/8/n/AC06565267/o/701010130063713873377220.jpg Qa. ∠BCD= (arc BC)/2 = ∠CAB ∠BDC= (arc BD)/2 = ∠DAB so, ∠CBD+∠CAD=∠CBD+∠CAB+∠DAB=∠CBD+∠BCD+∠BDC=180 Qb. ∠FBE+∠FAE=∠CBD+∠CAD=180 so, A,E,B,F are concyclic. Qc. ∠ADF=∠BDE ∠DAF=180-∠FBE=∠DBE so, △DAF~△DBE (AA) Qd. DA:BD=DF:DE (since △DAF~△DBE) 1+4 : 2 = 2+BF : 1 5=2(2+BF), so, BF= 1/2
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F5 Maths
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Figure: http://img291.imageshack.us/img291/1535/50487666.jpgIn Figure , two circles intersect each other at two points A and B. PQ is a common tangent to the two circles and touches them at points C and D. DB produced cuts AC at F and CB produced cuts AD at E.a) Prove that angle CBD + angle CAD =... 顯示更多 Figure: http://img291.imageshack.us/img291/1535/50487666.jpg In Figure , two circles intersect each other at two points A and B. PQ is a common tangent to the two circles and touches them at points C and D. DB produced cuts AC at F and CB produced cuts AD at E. a) Prove that angle CBD + angle CAD = 180 b) Prove that A, E, B and F are concyclic. c) Prove that triangle DAF ~ triangle DBE. d) If BD = 2, ED = 1 and AE = 4, find BF最佳解答:
圖片參考:http://imgcld.yimg.com/8/n/AC06565267/o/701010130063713873377220.jpg Qa. ∠BCD= (arc BC)/2 = ∠CAB ∠BDC= (arc BD)/2 = ∠DAB so, ∠CBD+∠CAD=∠CBD+∠CAB+∠DAB=∠CBD+∠BCD+∠BDC=180 Qb. ∠FBE+∠FAE=∠CBD+∠CAD=180 so, A,E,B,F are concyclic. Qc. ∠ADF=∠BDE ∠DAF=180-∠FBE=∠DBE so, △DAF~△DBE (AA) Qd. DA:BD=DF:DE (since △DAF~△DBE) 1+4 : 2 = 2+BF : 1 5=2(2+BF), so, BF= 1/2
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