F4 math problems ( Circles )

2007-07-19 1:33 pm
1. In the figure,triangleBCD ~ triangleEDA.
(a) Show thatA,B,D and E are concyclic points.
(b) Hence, deduce that BE//CD.
Figure http://server7.pictiger.com/img/422471/picture-hosting/--.jpg

2. In the figure, a circle with centre O is inscribed in a quadrilateral ABCD. AB⊥CB and AD⊥CD.AB = 12cm and CD = 4cm.
(a) Show that BC =DC.
(b) Show that OSBP is a square.
(c) Find the radius of the circle.
Figure http://server6.pictiger.com/img/422689/picture-hosting/4.php

回答 (1)

2007-07-19 2:12 pm
✔ 最佳答案

圖片參考:http://i182.photobucket.com/albums/x4/A_Hepburn_1990/circle1-1.jpg?t=1184795529


1. a. Let ∠BDC be x, ∠BCD be y

So, ∠ABD = x + y (ext. ∠ of Δ)

Since ΔBCD ~ ΔEAD

So, ∠AED = ∠BCD = 180° - x – y (corr. ∠s of ~Δs)

∠AED + ∠ABD
= (180° - x – y) + (x + y)
= 180°

So, A, B, D, E are concyclic. (opp. ∠s eq.)



b. Since A, B, D, E are concyclic,

So, ∠DBE = ∠DAE (∠s in the same segment)

Since ΔBCD ~ ΔEAD

So, ∠BDC = ∠DAE (corr. ∠s of ~Δs )

So, ∠BDC = ∠DBE

So, BE // CD (alt. ∠s equal)





圖片參考:http://i182.photobucket.com/albums/x4/A_Hepburn_1990/circle2.jpg?t=1184796007




2. ∠ABC = ∠ADC = 90° (given)
∠ACB = ∠ACD (tangent property)
AC = AC (common)

So, ΔABC is congruent to ΔADC (AAS).

So, BC = DC (corr. sides of con. Δs)
∠OPB = ∠OSB = 90° (tangent ⊥ radius)
So, ∠POS = 90°

OS = OP (radius)

So, OSBP is a square.

c. AC = √(AB2 + BC2)
AC = √(122 + 42)
AC = 4√10 cm

Let r cm be the radius of the circle.

∠OSC = ∠ABC = 90°
∠OCS = ∠ACB (common)

So, ΔABC ~ ΔOSC (AA)

So, AB / OS = BC / SC = AC / OC

AB / OS = BC / SC

12 / r = 4 / (4 - r)

48 – 12r = 4r

r = 3

So, the radius of the circle is 3 cm.

b.
參考: Myself~~~


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