As shown in the figure, in the square ABCD, e is the midpoint of AD, f is on CD, and DF = 1 / 4CD BE.EF.BF To prove that be is perpendicular to ef
Let X be the side length of a square
Ed = x / 2, DF = x / 4. According to Pythagorean theorem, EF = radical 5 / 4x,
In the same way,
Be = root 5 / 2x, BF = 5 / 4x,
Then be square + EF square = be square
So the triangle bef is a right triangle
So be vertical EF
(for reference only)
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