※ 引述《Intercome (今天的我小帥)》之銘言:
: △ABC,BA = 15,在AC邊上找D、E兩點,使得AE=5、DE=11、CD=9
: 且∠CBD=∠EBA,則BE=________
依據wayn2008所提供的提示我整理解題過程如下,再請各位參考~~
首先設BC = x, BD = y, BE = a
由△ABD:△BCE = 16:20 = 4:5 = 15y:ax => 75y = 4ax
△ABE:△BCD = 5:9 = 15a:xy => xy = 27a
因為∠CBD=∠EBA 由餘弦定理
225+a^2-25 x^2+y^2-81 x^2+y^2-81
=> ----------- = ---------- = ---------- => 1800+9a^2 = 5x^2+5y^2-405 ..(1)
2*a*15 2xy 2*27a
又因為 ∠CBE=∠DBA 由餘弦定理
x^2+a^2-400 225+y^2-256 x^2+a^2-400
=> ----------- = ------------ = ------------ => 5y^2 = 4x^2+4a^2-1445 ..(2)
2ax 2*15y 2*75y/4
將(2)代入(1): 5a^2 = 9x^2-3650 ..(3)
最後因為 ∠BAE=∠BAC 由餘弦定理
225+25-a^2 225+625-x^2
=> ---------- = ----------- => 1250-5a^2 = 850-x^2 將(3)代入
2*5*15 2*15*25
=> 1250 - 9x^2 + 3650 = 850 - x^2, x^2 = 4050/4 代回(3)
5a^2 = 18225/4 - 14600/4 = 3625/4, a = 5√29/2 #
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