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發問:
圖片參考：http://imgcld.yimg.com/8/n/HA00517423/o/701207290052413873413280.jpg in the figure,o is the centre of the circle.AC and BD are chords intersecting at E such that BE=CE.1.SHOW THAT A,E ,O AND B ARE CONCYCLIC.2.IF ANGLE OEB=56 DEGREE, FIND ANGLE AEB AND ANGLE ODC.ANSWER:ANGLE AEB=68 DEGREE, ANGLE ODC=56... 顯示更多 圖片參考：http://imgcld.yimg.com/8/n/HA00517423/o/701207290052413873413280.jpg in the figure,o is the centre of the circle.AC and BD are chords intersecting at E such that BE=CE. 1.SHOW THAT A,E ,O AND B ARE CONCYCLIC. 2.IF ANGLE OEB=56 DEGREE, FIND ANGLE AEB AND ANGLE ODC. ANSWER:ANGLE AEB=68 DEGREE, ANGLE ODC=56 DEGREE
最佳解答:
1) ∠CAB = (1/2)∠COB (∠at centre twice ∠ at ☉ce) Let M be the mid point of BC , then OM ⊥ BC and EM ⊥ BC since EM is the height of isos.△ BEC , so EOM is a straight line So ∠BOM = (1/2)∠COB =∠CAB , therefore A,E ,O AND B ARE CONCYCLIC. (ext.∠= int. opp.∠) 2) ∠EBC =∠ECB = 90° - ∠OEB = 90 - 56 = 34° so∠AEB = ∠EBC + ∠ECB = 34° + 34°= 68° (ext. ∠ of Δ)∠ODC = (180° - ∠DOC)/2 = (180° - 2∠EBC )/2 (∠at centre twice ∠ at ☉ce) = (180° - 2*34° )/2 = 56°
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maths question發問:
圖片參考：http://imgcld.yimg.com/8/n/HA00517423/o/701207290052413873413280.jpg in the figure,o is the centre of the circle.AC and BD are chords intersecting at E such that BE=CE.1.SHOW THAT A,E ,O AND B ARE CONCYCLIC.2.IF ANGLE OEB=56 DEGREE, FIND ANGLE AEB AND ANGLE ODC.ANSWER:ANGLE AEB=68 DEGREE, ANGLE ODC=56... 顯示更多 圖片參考：http://imgcld.yimg.com/8/n/HA00517423/o/701207290052413873413280.jpg in the figure,o is the centre of the circle.AC and BD are chords intersecting at E such that BE=CE. 1.SHOW THAT A,E ,O AND B ARE CONCYCLIC. 2.IF ANGLE OEB=56 DEGREE, FIND ANGLE AEB AND ANGLE ODC. ANSWER:ANGLE AEB=68 DEGREE, ANGLE ODC=56 DEGREE
最佳解答:
1) ∠CAB = (1/2)∠COB (∠at centre twice ∠ at ☉ce) Let M be the mid point of BC , then OM ⊥ BC and EM ⊥ BC since EM is the height of isos.△ BEC , so EOM is a straight line So ∠BOM = (1/2)∠COB =∠CAB , therefore A,E ,O AND B ARE CONCYCLIC. (ext.∠= int. opp.∠) 2) ∠EBC =∠ECB = 90° - ∠OEB = 90 - 56 = 34° so∠AEB = ∠EBC + ∠ECB = 34° + 34°= 68° (ext. ∠ of Δ)∠ODC = (180° - ∠DOC)/2 = (180° - 2∠EBC )/2 (∠at centre twice ∠ at ☉ce) = (180° - 2*34° )/2 = 56°
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