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標題:
circle
發問:
1)ABCD is a quadrilateral inscribed in a circle. If angle ADB = 28° and angle CBD = 62°, prove that arc AB + arc CD = arc BC + arc AD. (no diagram is given)我只係知angle ADB + angle CBD = angle ABD + angle BDC = 90° (opp. angles, cyclic quad.) 但係arc AB + arc CD =/= angle ADB + angle CBD 咁點算???... 顯示更多 1) ABCD is a quadrilateral inscribed in a circle. If angle ADB = 28° and angle CBD = 62°, prove that arc AB + arc CD = arc BC + arc AD. (no diagram is given) 我只係知angle ADB + angle CBD = angle ABD + angle BDC = 90° (opp. angles, cyclic quad.) 但係arc AB + arc CD =/= angle ADB + angle CBD 咁點算??? 要let數??? 2) The diagonals of a cyclic quadrilateral ABCD intersect at K. DA, CB when produced meet at E; AB, DC when produced meet at F. If angle AEB = 55°, angle BFC = 35° and angle DKC = 85°, prove that arc BC is twice arc AB. (no diagram is given) angle CAB and angle ACB 唔識搵,計到勁多angle都係xx.5° 更新: 題目冇比圖,係咪要自己估個圖係點?有冇人係唔駛畫返都識做=.=?
最佳解答:
1) Let O be the centre of the circle as follows: 圖片參考：http://i1191.photobucket.com/albums/z467/robert1973/Nov12/Crazygeom2.jpg Let ∠ADB = θ and ∠CDB = α, then∠DAK = 85 - θ (Ext, angle of triangle) ∠DCK = 95 - α (Ext, angle of triangle)∠ACB = θ and ∠CAB = α (Angles in the same segment)∠DAK = ∠AEC + ∠ACB (Ext. angle of triangle)85 - θ = 55 + θθ = 15 ∠DCK = ∠AFC + ∠CAB (Ext. angle of triangle) 95 - α = 35 + αα = 30With α = 2θ, arc BC = 2 x arc AB
其他解答:
circle
發問:
1)ABCD is a quadrilateral inscribed in a circle. If angle ADB = 28° and angle CBD = 62°, prove that arc AB + arc CD = arc BC + arc AD. (no diagram is given)我只係知angle ADB + angle CBD = angle ABD + angle BDC = 90° (opp. angles, cyclic quad.) 但係arc AB + arc CD =/= angle ADB + angle CBD 咁點算???... 顯示更多 1) ABCD is a quadrilateral inscribed in a circle. If angle ADB = 28° and angle CBD = 62°, prove that arc AB + arc CD = arc BC + arc AD. (no diagram is given) 我只係知angle ADB + angle CBD = angle ABD + angle BDC = 90° (opp. angles, cyclic quad.) 但係arc AB + arc CD =/= angle ADB + angle CBD 咁點算??? 要let數??? 2) The diagonals of a cyclic quadrilateral ABCD intersect at K. DA, CB when produced meet at E; AB, DC when produced meet at F. If angle AEB = 55°, angle BFC = 35° and angle DKC = 85°, prove that arc BC is twice arc AB. (no diagram is given) angle CAB and angle ACB 唔識搵,計到勁多angle都係xx.5° 更新: 題目冇比圖,係咪要自己估個圖係點?有冇人係唔駛畫返都識做=.=?
最佳解答:
1) Let O be the centre of the circle as follows: 圖片參考：http://i1191.photobucket.com/albums/z467/robert1973/Nov12/Crazygeom2.jpg Let ∠ADB = θ and ∠CDB = α, then∠DAK = 85 - θ (Ext, angle of triangle) ∠DCK = 95 - α (Ext, angle of triangle)∠ACB = θ and ∠CAB = α (Angles in the same segment)∠DAK = ∠AEC + ∠ACB (Ext. angle of triangle)85 - θ = 55 + θθ = 15 ∠DCK = ∠AFC + ∠CAB (Ext. angle of triangle) 95 - α = 35 + αα = 30With α = 2θ, arc BC = 2 x arc AB
其他解答:
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