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F.4 more abour circles

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1)http://i729.photobucket.com/albums/ww291/kaka1307/DSC00551.jpgAN inscribed circle touches the three sides AB,BC and CA of triangle ABCat three points Z,X and Y respectively.If angle ABC=52 and angle ZXY=44find a)angle ACBb)XYZ2)http://i729.photobucket.com/albums/ww291/kaka1307/DSC00553.jpgBA and DF are... 顯示更多 1) http://i729.photobucket.com/albums/ww291/kaka1307/DSC00551.jpg AN inscribed circle touches the three sides AB,BC and CA of triangle ABC at three points Z,X and Y respectively.If angle ABC=52 and angle ZXY=44 find a)angle ACB b)XYZ 2) http://i729.photobucket.com/albums/ww291/kaka1307/DSC00553.jpg BA and DF are two parallel tangents which touch the circle at two ponits B and D respectively. AE is the tangent to the circle at ponit E and AE is produced to meet DF at point C.If angle ACF=70 find a)angle OAE b)angle AOC 3) http://i729.photobucket.com/albums/ww291/kaka1307/DSC00555.jpg The 4 sides of the quadrilateral ABCD are the tangents to the circle at P,Q,R and S respectively.If AB=26 and DC=12 a)determine whether AD+BC is equal to AB+DC b)find the perimeter of the quadrilateral answer: 1a)36 b)64 2a)35 b)90 3a)yes b)76 更新: Plese shoe clear steps.thx! 更新 2: Please

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1a) By tangent properties, BZ = BX and CY = CX and hence △BXZ and △CXY are isos. △. So ∠BXZ = ∠BZX = (180 - 52)/2 = 64 ∠CXY = 180 - 64 - 44 = 72 ∠CYX = 72 ∠ACB = 180 - 72 - 72 = 36 b) ∠XZY = ∠CXY = 72 (∠ in alt. segment) ∠XYZ = 180 - 44 - 72 = 64 2a) By tangent properties, OA is the angle bisector of ∠BAE Also ∠BAE = ∠ACF = 70 (Alt. ∠, AB//CF) ∠OAE = 35 b) Similarly, OC is the angle bisector of ∠DCE, thus ∠OCE = 110/2 = 55 So ∠AOC = 180 - 35 - 55 = 90 3a) By tangent properties, DS = DR, CR = CQ, AS = AP and BP = BQ So AS + DS + BQ + CQ = AP + BP + DR + CR AD + BC = AB + DC b) Perimeter = AD + BC + AB + DC = 2(AB + DC) = 76

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