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Mathematics Competition 1958
Are the conditions sufficient? In triangle ABC, AD is the height of BC, P is any point on AD, and the extension lines of BP and CP are AC at E and AB at F respectively. Proof: ∠ADE=∠ADF.

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.