✔ 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.