If f is any point on BC of square ABCD, AE bisects ∠ DAF intersecting CD and E, the proof is: AF = BF + De Can you get full marks in the senior high school entrance examination? You are only suitable for the analysis of filling in the blanks, but you will never get full marks for the answers. Now I'm sitting on the chair. What you did is wrong!

If f is any point on BC of square ABCD, AE bisects ∠ DAF intersecting CD and E, the proof is: AF = BF + De
Because f is any point of the edge BC, we can extremely Let f and C coincide, and let the side length of the square be 1,
Then AF = AC = √ 2, BF = BC = 1
DE=1*tg(45/2)=√2-1
So there is: BF + de = 1 + √ 2-1 = √ 2
That is AF = BF + De

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