急! Properties of Circles 12q15

2012-03-03 6:14 pm
請詳細步驟教我計以下四條 :
(不要綱址回答)


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回答 (1)

2012-03-03 9:53 pm
✔ 最佳答案
34)(64 - 25)² + ?² = 95²
?² = 95² - 39²
? = √7504
? = 87(nearest integer)
1)∠ONB = 90° (line joining centre to mid-pt. of chord⊥chord)
∠CBN = 90° (property of rectangle)
∴ BC // ON (alt.∠s equal)∠BCP = ∠CBN = 90° (property of rectangle)
∴ CP // BN (int.∠s supp.)
∴ ∠CPN = ∠BNO = 90° (corr.∠s, CP // BN)
CPD is a tangent to circle O. (converse of tangent⊥radius)
∴CPD touches the circle at P.
2)Join OO' , meeting AB at C.∠OCA =∠O'CB (vert. opp. ∠s)
∠COA = ∠CO'B (alt.∠s, OA // BO')
∴ ΔCOA ~ ΔCO'BSo ∠O'BA = ∠OAB = 90° (tangent⊥radius)
∴ AB is a tangent to circle O' (converse of tangent⊥radius)
6)∠CAB = ∠CAB (common)
AC : AB = 1 : 1 (given)
AE : AD = 1 : 1 (given)
∴ ΔABC ~ ΔADE (S.A.S.)So ∠CED = ∠ACB = ∠ABC (base ∠s, isos.Δ)∴ BCDE is a cyclic quadrilateral. (ext.∠= int. opp.∠)


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