
Re: [中學] 圓


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: Sent from JPTT on my Asus ASUS_Z01RD.
這題很繁
∠CDO' = a = ∠O'CD
=> ∠CO'O = 2a = 90 - ∠O'OC = 90 - 2∠DBO
=> ∠DBO = 45 - a = ∠BDO
=> ∠BDC = 45 = ∠ADC
又因為∠DCA = ∠DEC
=> △DCA ~ △DEC
DA : DB = 2 : 3
(2k)^2 + (3k)^2 = 25^2 => k = 25/√13 => 求出DA, DB
DA * DB = (DA + DB) * CD/√2 => 可求出CD
CE : 15 = CD : BD
=> CE = 15 CD / BD 可算出一個答案
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