P is any point on the side BC of the square ABCD, and PE is perpendicular to BD intersecting BDE, PF is perpendicular to AC intersecting AC, if AC = 10, EP = FP

Should it be "EP + FP"?
Let AC and BD intersect o
Because the quadrilateral ABCD is a square
So ob = OC = AC / 2 = 5, ob ⊥ OC
Because PE ⊥ ob, PF ⊥ OC
So the quadrilateral peof is a rectangle
So EP = of
The triangle PCF is an isosceles right triangle
So FP = CF
So EP + FP = of + CF = OC
So EP + FP = 5
For reference! Jswyc

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