If the bisectors of parallelogram ABCD, ab = 5, ad = 8, angles C and D intersect ad, BC and points E and f respectively, and AF is perpendicular to BC, the solution of CE is obtained

If the bisectors of parallelogram ABCD, ab = 5, ad = 8, angles C and D intersect ad, BC and points E and f respectively, and AF is perpendicular to BC, the solution of CE is obtained

[I've thought about it for a long time. It's hand-made and pollution-free. I hope to adopt it]
EC and DF intersect at point O and connect EF
In the parallelogram ABCD,
∵AD∥BC
The bisectors of angles C and D intersect ad, BC and points E and f respectively
∴DF⊥EC
In △ Dec,
∵∠FDC=∠FDE,∠FDC+∠DCE=∠FDE+∠DEC=90°
The result shows that the delta Dec is an isosceles triangle, that is, de = DC = 5
In △ CDF,
Similarly, we get that △ CDF is an isosceles triangle, that is FC = DC = 5
Also ∵ de ∥ CF and de = CF (opposite sides parallel and equal)
The quadrilateral DEFC is a parallelogram
And ∵ de = DC = CF = 5 (a group of parallelograms with equal edges)
The parallelogram DEFC is a quadrilateral
[in this way, the length of most edges is determined]
AE=BF=3,EF=AB=5
⊙ AF ⊥ BC (second known condition)
∴AF=4
In △ ADF, ∠ fad = 90 °
It is concluded that DF = 4 √ 5
In a quadrilateral DEFC, the diagonal is vertical and bisected
That is: do = 2 √ 5
In △ deo
∵EO⊥DO,DE=5
∴EO=√5
EO=OC
∴CE=2√5
As long as the proof of efcd of quadrilateral is completed, the problem will be simple_ ∩) O ha ha ~]