標題:
F.5 circles (20marks)!!!!!!!!!
發問:
1.The figure shows two intersecting equal circles with centres O and O' respectively.It is given that AOB, BCD and AO'D are straight lines.Show that ΔCDO' is an equilateral triangle.2.In the figure, AB is a diameter of the circle. M is an external point of the circle such that MO⊥AB. MB intersects the... 顯示更多 1.The figure shows two intersecting equal circles with centres O and O' respectively.It is given that AOB, BCD and AO'D are straight lines.Show that ΔCDO' is an equilateral triangle. 2.In the figure, AB is a diameter of the circle. M is an external point of the circle such that MO⊥AB. MB intersects the circle at P. AP intersects MO at N.Prove that (a) O,P,M and A are concyclic, (b) ∠OPA = ∠OMB. 3.In the figure, ABis a diameter of the circle. O is a point on AB such that ∠APQ = ∠ARP. AR intersects PQ at T. Prove that (a) ∠PQB = 90° (b) R, T, Q and B are concyclic. (請詳列計算過程，圖在下) 圖片參考：http://imgcld.yimg.com/8/n/HA00763813/o/701012010080513873419070.jpg
最佳解答:
1. Since OA = OC = OO' = O'A = O'C = O'D (radius) Thus triangle OAO' is equilateral and OAO'C is a rhombus. Angle OAO' = 60 OA//CO' (property of rhombus) Hence CO'D = 60 (corr. angles, OA//O'C) Since angle O'CD = angle O'DC (base angles, isos. triangle) So 180 - 2(angleO'CD) = 60 angleO'CD = 60 = angleO'DC So triangle CDO' is equilateral. 2a. Consider triangle ABN and triangle MPN, AP perpendicular BM (angle in semi-circle) Hence angleAOM = angleAPM = 90 Thus O,P,M,A are concyclic. (converse of angles in the same segment) b. Note that OA = OP (radius) Thus by (a), angleAPO = angleOMP (equal chords, equal angles) 3a. Join PB. Let angleAPQ = a anglePBA = anglePRA = a (angles in the same segment) angleAPB = 90 (angle in semi-circle) anglePAB = 180 - 90 - a (angle sum of triangle) = 90 - a So anglePQA = 180 - a - (90 - a) (angle sum of triangle) = 90 Thus angle PQB = 180 - 90 (adj. angles on st. line) = 90 b. angleARB = 90 (angle in semi-circle) Hence angleARB + anglePQB = 90 + 90 = 180 So R,T,Q,B are concyclic. (opp. angles supp.) 2010-12-01 18:13:06 補充： 1. Since angle O'CD = angle O'DC (base angles, isos. triangle) So 180 - 2(angleO'CD) = 60 (angle sum of triangle)
其他解答:
F.5 circles (20marks)!!!!!!!!!
發問:
1.The figure shows two intersecting equal circles with centres O and O' respectively.It is given that AOB, BCD and AO'D are straight lines.Show that ΔCDO' is an equilateral triangle.2.In the figure, AB is a diameter of the circle. M is an external point of the circle such that MO⊥AB. MB intersects the... 顯示更多 1.The figure shows two intersecting equal circles with centres O and O' respectively.It is given that AOB, BCD and AO'D are straight lines.Show that ΔCDO' is an equilateral triangle. 2.In the figure, AB is a diameter of the circle. M is an external point of the circle such that MO⊥AB. MB intersects the circle at P. AP intersects MO at N.Prove that (a) O,P,M and A are concyclic, (b) ∠OPA = ∠OMB. 3.In the figure, ABis a diameter of the circle. O is a point on AB such that ∠APQ = ∠ARP. AR intersects PQ at T. Prove that (a) ∠PQB = 90° (b) R, T, Q and B are concyclic. (請詳列計算過程，圖在下) 圖片參考：http://imgcld.yimg.com/8/n/HA00763813/o/701012010080513873419070.jpg
最佳解答:
1. Since OA = OC = OO' = O'A = O'C = O'D (radius) Thus triangle OAO' is equilateral and OAO'C is a rhombus. Angle OAO' = 60 OA//CO' (property of rhombus) Hence CO'D = 60 (corr. angles, OA//O'C) Since angle O'CD = angle O'DC (base angles, isos. triangle) So 180 - 2(angleO'CD) = 60 angleO'CD = 60 = angleO'DC So triangle CDO' is equilateral. 2a. Consider triangle ABN and triangle MPN, AP perpendicular BM (angle in semi-circle) Hence angleAOM = angleAPM = 90 Thus O,P,M,A are concyclic. (converse of angles in the same segment) b. Note that OA = OP (radius) Thus by (a), angleAPO = angleOMP (equal chords, equal angles) 3a. Join PB. Let angleAPQ = a anglePBA = anglePRA = a (angles in the same segment) angleAPB = 90 (angle in semi-circle) anglePAB = 180 - 90 - a (angle sum of triangle) = 90 - a So anglePQA = 180 - a - (90 - a) (angle sum of triangle) = 90 Thus angle PQB = 180 - 90 (adj. angles on st. line) = 90 b. angleARB = 90 (angle in semi-circle) Hence angleARB + anglePQB = 90 + 90 = 180 So R,T,Q,B are concyclic. (opp. angles supp.) 2010-12-01 18:13:06 補充： 1. Since angle O'CD = angle O'DC (base angles, isos. triangle) So 180 - 2(angleO'CD) = 60 (angle sum of triangle)
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