Next, I will use the plane geometry of junior high school to prove it. I don't need the Ceva theorem that junior high school students haven't learned, but only the similar triangles property that junior high school students have learned.
Prove:
Pass a as LK//BC, and extend the intersection of CF, DF, DE and BE at L, H, G and K respectively.
According to the nature of similar triangles.
AG/CD=AK/BC,AG=AK*CD/BC - ( 1)
AH/BD=AL/BC,AH=BD*AL/BC - (2)
In the same way.
AK/BD=AL/CD,AK*CD=BD*AL - (3)
From (1)(2)(3)
AG=AH,
Because DA is the height on the triangle DGH.
So: ∠ADG=∠ADH (isosceles triangle three-line unity theorem)
That is ∠ADE=∠ADF.
I really don't know. The question is:
"The third (1993) Macau Mathematics Competition;
14th (200 1) Irish Mathematics Competition;
18th (1958) Putnam Mathematics Competition;
26th (1994) Canadian Mathematics Competition;
The champion of the first (1987) "Friendship Cup" international mathematics competition.