Question

一道几何解答题

Answer 1

(1) 设DC=AD=BD=1,则AB=2cosA，BC=2sinA
ED⊥AC=>EC=CD/cosC=1/sinA
=>EO=OC=1/(2sinA)，BE=2sinA-1/sinA
由BF·BD=BE·BC=>BF=4sin²A-2
∵OF//AC,∴BF/BD=BO/BC
=>4sin²A-2=(2sinA-1/(2sinA))/(2sinA)
令x=4sin²A,则上式变为
x-2=(x-1)/x => x²-3x+1=0
=>x=(3±√5)/2。又∵sinA=BC/2≥EC/2=CD/2=1/2
∴x=4sin²A>1，∴x=(3+√5)/2
=>cos²A=(4-x)/4=(5-√5)/8=>cosA=[√(10-2√5)]/4
(2) F是BD中点，则BF=1/2，由上可知
即4sin²A-2=1/2=>sin²A=5/8
=>cos²A=3/8 =>cosA=√6/4