F.5 circles (20marks)!!!!!!!!!

chi ming

2010-12-02 1:43 am

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

回答 (1)

蛙人

2010-12-02 2:11 am

✔ 最佳答案

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)

參考: me

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