Question

初中数学几何证明

已知：BD、CE是△ABC的角平分线且相较于点O，OF⊥ED交BC于点G，OG=2OF。求证：∠A=60°

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Answer 1

设内切圆半径1
角BAC＝2x
角ABC＝2y
角BCA＝2z
角EOF＝a
角FOD＝b
a+b=90+x
OG=g
OF=f

那么
OB＝1/siny
OC=1/sinz
OE=1/sin(2y+z)
OD=1/sin(2z+y)

x+y+z=90
a+b=90+x
cosa/sin(2y+z)=f=cosb/sin(2z+y)
对三角形OGC，用正弦定理
g/sinz=(1/sinz)(1/sin(a+z))
所以g=1/sin(a+z)

已知OG=2OF要证x＝30度
f=cosa/sin(2y+z)

f=cos(90+x-a)/sin(2z+y)
g=1/sin(a+z)

已知OG=2OF要证x＝30度

f=cosa/sin(2y+z)

f=cos(90+x-a)/sin(2z+y)

g=1/sin(a+z)

已知g=2f

2f=2cosa/sin(2y+z)

2f= -2cos(y+z+a)/sin(2z+y)

g=1/sin(a+z)

要证y+z＝60度

已知

2cosa/sin(2y+z)= -2cos(y+z+a)/sin(2z+y) =1/sin(a+z)

要证y+z＝60度

2sin(a+z)cosa=sin(2y+z)

2sin(a+z)cos(y+z+a)=-sin(2z+y)

!!!!!!!!!!! sin(A+B)+sin(A-B)=2sinAcosB

sin(2a+z)+sin(z)=sin(2y+z)
sin(y)=sin(2z+y)+sin(2a+y+2z)
等价于
sin(2a+z)=sin(2y+z)-sinz
sin(2a+z+y+z)=sin(y)-sin(2z+y)

!!!!!!!!!!! sin(A+B)-sin(A-B)=2cosAsinB

等价于
sin(2a+z)=sin(2y+z)-sinz=2cos(y+z)sin(y)
sin(2a+z+y+z)=sin(y)-sin(2z+y)=2cos(y+z)sin(-z)

等价于
sin(2a+z)=2cos(y+z)sin(y)
sin(2a+z+y+z)=2cos(y+z)(-sinz)

等价于
sin(2a+z+y+z)sin(y)=sin(2a+z)(-sinz)
sin(2a+z)=2cos(y+z)sin(y)

等价于
sin(2a+z)=2cos(y+z)sin(y)

sin(y)[sin(2a+z)cos(y+z)+sin(y+z)cos(2a+z)]=-sin(2a+z)sin(z)

[sin(y)sin(y+z)]cos(2a+z)=sin(2a+z)[-sin(z)-sin(y)cos(y+z)]

等价于

sin(2a+z)=2cos(y+z)sin(y)
[sin(y)sin(y+z)]²cos(2a+z)²=sin(2a+z)²[sin(z)+sin(y)cos(y+z)]²

等价于
sin(2a+z)=2cos(y+z)sin(y)
[sin(y)sin(y+z)]²[1-sin(2a+z)²]=sin(2a+z)²[sin(z)+sin(y)cos(y+z)]²

等价于
sin(2a+z)=2cos(y+z)sin(y)
[sin(y)sin(y+z)]²=sin(2a+z)²{[sin(z)+sin(y)cos(y+z)]²+[sin(y)sin(y+z)]²}

等价于
sin(2a+z)=2cos(y+z)sin(y)
[sin(y)sin(y+z)]²=[2cos(y+z)sin(y)]²{[sin(z)+sin(y)cos(y+z)]²+[sin(y)sin(y+z)]²}

等价于
sin(2a+z)=2cos(y+z)sin(y)
[tan(y+z)]²=4{[sin(z)+sin(y)cos(y+z)]²-[sin(y)sin(y+z)]²}

等价于
sin(2a+z)=2cos(y+z)sin(y)

(1/4)[tan(y+z)]²=sin(z)²+sin²(y)cos²(y+z)+2sin(y)sin(z)cos(y+z)+sin²(y)sin²(y+z)

(1/4)[tan(y+z)]²=sin(z)²+sin²(y)+2sin(y)sin(z)cos(y+z)

(1/2)[tan(y+z)]²+1-2sin(z)²+1-2sin²(y)=2+4sin(y)sin(z)cos(y+z)
(1/2)[tan(y+z)]²+cos2z+cos2y=2+4sin(y)sin(z)cos(y+z)
cos(A+B)=cosAcosB-sinAsinB
cos(A-B)=cosAcosB+sinAsinB
2sinAsinB=cos(A-B)-cos(A+B)
(1/2)[tan(y+z)]²+2cos(z+y)cos(z-y)=2+2cos(y+z)[cos(y-z)-cos(y+z)]

(1/2)[tan(y+z)]²=2-2cos²(y+z)

sin²(y+z)=[4-4cos²(y+z)]cos²(y+z)
1-cos²(y+z)=[4-4cos²(y+z)]cos²(y+z)
设cos²(y+z)=t

1-t=(4-4t)t
4t²-5t+1=0
(4t-1)(t-1)=0
t=1/4或t=1(舍去)
cos(y+z)=1/2或-1/2(舍去)
0<y+z<90度
y+z=60度
x=30度
角A＝2x＝60度

Answer 2

∵BD、CE是△ABC的角平分线且相较于点O
∴BO=2OD
又∵OG=2OF
∠BOG=∠DOF
∴△BOG∽△DOF
同理△COG∽△EOF
则∠OGB=90°
OG=1/2OB=1/2CO
∴∠ABC=∠ACB=60°
∴∠A=60°

追问

看这个图 那个图太特殊

Answer 3

初几的题？

追问

初中的