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MATH QUESTIONS 2 [CIRCLES]
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34. The figure shows a hexagon ABCDEF inscribed in a circle. angle FAB =x, angle BCD = y and angle DEF = z. Find x+y+z. 圖片參考：http://imgcld.yimg.com/8/n/HA00610265/o/701009270030413873373900.jpg 36. In the figure, circle ABD passes through the centre O of circle BCD. IF angle BCD =70, then angle DAB... 顯示更多 34. The figure shows a hexagon ABCDEF inscribed in a circle. angle FAB =x, angle BCD = y and angle DEF = z. Find x+y+z. 圖片參考：http://imgcld.yimg.com/8/n/HA00610265/o/701009270030413873373900.jpg 36. In the figure, circle ABD passes through the centre O of circle BCD. IF angle BCD =70, then angle DAB = 圖片參考：http://imgcld.yimg.com/8/n/HA00610265/o/701009270030413873373901.jpg 37. In the figure, BCE is a straight line. AD = BD. Which of the following must be correct? 圖片參考：http://imgcld.yimg.com/8/n/HA00610265/o/701009270030413873373912.jpg I. DC is an angle bisector of angle ACE. II. angle DBA + angle ACB = 90 III. Tirangle ABF ~ Triangle DCF THX SO MUCH!
34. Join FC ABCF and CDEF are cyclic quad. ∠FAB+∠BCF=180? (opp. ∠s, cyclic quad.) ∠DCF+∠DEF=180? ∠FAB+∠BCF+∠DCF+∠DEF=360? ∴ x+y+z=360? 36. Join BO and DO In circle BCD, ∠BOD=2∠BCD=140? (∠ at center = twice ∠ at circumference) In circle ABD, ABOD is a cyclic quad. ∠DAB+∠BOD=180? (opp. ∠s, cyclic quad.) ∠DAB+140?=180? ∴ ∠DAB=40? 37. ∠ACD=∠ABD (∠s in the same segment) ∠DCE=∠BAD (ext. ∠, cyclic quad) ∠ABD=∠BAD (base∠s, isos. △) ∠ACD=∠DCE DC is an angle bisector of angle ACE. ∴ I. must be correct ∠ACB=∠ADB (∠s in the same segment) ∠DBA+∠ACB=∠DBA+∠ADB ∠DBA+∠ADB+∠BAD=180? (∠s sum of △) ∠DBA+∠ADB= =180?-∠BAD ∠BAD and ∠ABD are base ∠s of isos. △ABD ∠BAD=∠ABD<90? ∠DBA+∠ACB>90? ∴ II. is NOT correct Assume that AC and BD intersect at F ∠ABF=∠FCD (∠s in the same segment) ∠BAF=∠FDC (∠s in the same segment) △ABF ~ △DCF (AAA) ∴ III. must be correct
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MATH QUESTIONS 2 [CIRCLES]
發問:
34. The figure shows a hexagon ABCDEF inscribed in a circle. angle FAB =x, angle BCD = y and angle DEF = z. Find x+y+z. 圖片參考：http://imgcld.yimg.com/8/n/HA00610265/o/701009270030413873373900.jpg 36. In the figure, circle ABD passes through the centre O of circle BCD. IF angle BCD =70, then angle DAB... 顯示更多 34. The figure shows a hexagon ABCDEF inscribed in a circle. angle FAB =x, angle BCD = y and angle DEF = z. Find x+y+z. 圖片參考：http://imgcld.yimg.com/8/n/HA00610265/o/701009270030413873373900.jpg 36. In the figure, circle ABD passes through the centre O of circle BCD. IF angle BCD =70, then angle DAB = 圖片參考：http://imgcld.yimg.com/8/n/HA00610265/o/701009270030413873373901.jpg 37. In the figure, BCE is a straight line. AD = BD. Which of the following must be correct? 圖片參考：http://imgcld.yimg.com/8/n/HA00610265/o/701009270030413873373912.jpg I. DC is an angle bisector of angle ACE. II. angle DBA + angle ACB = 90 III. Tirangle ABF ~ Triangle DCF THX SO MUCH!
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最佳解答:34. Join FC ABCF and CDEF are cyclic quad. ∠FAB+∠BCF=180? (opp. ∠s, cyclic quad.) ∠DCF+∠DEF=180? ∠FAB+∠BCF+∠DCF+∠DEF=360? ∴ x+y+z=360? 36. Join BO and DO In circle BCD, ∠BOD=2∠BCD=140? (∠ at center = twice ∠ at circumference) In circle ABD, ABOD is a cyclic quad. ∠DAB+∠BOD=180? (opp. ∠s, cyclic quad.) ∠DAB+140?=180? ∴ ∠DAB=40? 37. ∠ACD=∠ABD (∠s in the same segment) ∠DCE=∠BAD (ext. ∠, cyclic quad) ∠ABD=∠BAD (base∠s, isos. △) ∠ACD=∠DCE DC is an angle bisector of angle ACE. ∴ I. must be correct ∠ACB=∠ADB (∠s in the same segment) ∠DBA+∠ACB=∠DBA+∠ADB ∠DBA+∠ADB+∠BAD=180? (∠s sum of △) ∠DBA+∠ADB= =180?-∠BAD ∠BAD and ∠ABD are base ∠s of isos. △ABD ∠BAD=∠ABD<90? ∠DBA+∠ACB>90? ∴ II. is NOT correct Assume that AC and BD intersect at F ∠ABF=∠FCD (∠s in the same segment) ∠BAF=∠FDC (∠s in the same segment) △ABF ~ △DCF (AAA) ∴ III. must be correct
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