RT △ ABC, ∠ a = 90 °, the bisector of ∠ B intersects AC at D, the vertical line from a as BC intersects BD at e, and the bisector from D as DF ⊥ BC

RT △ ABC, ∠ a = 90 °, the bisector of ∠ B intersects AC at D, the vertical line from a as BC intersects BD at e, and the bisector from D as DF ⊥ BC

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Rectangle cdef is RT triangle ABC inscribed rectangle, ∠ C = 90 ° BC-AC = 2, let ad = x, cdef area is y, rectangle maximum area is S1, triangle area is s
Find S1 / S

1/2

In RT △ ABC, ∠ C = 90 °, OEF is the inscribed square of triangle, and BD intersects EF with O, AC = 2, ab = 2 root sign 17. Find the area of square cdef
CB = 8. I see
The next step is the whole process

Let the side length of a square cdef be X,
According to the similarity between RT △ ABC and RT △ ADF, AD / AC = DF / BC, i.e. (2-x) / 2 = x / 8
So x = 1.6,
The square cdef area is 2.56