Just do the following questions, because the front is relatively simple:

(2) ① Verification: CE= 1/2AB.

It is proved that extending CE to F, making EF=CE and connecting A'F and B'F, then quadrilateral A'CB'F is a parallelogram (quadrilaterals whose diagonals bisect each other are parallelograms) ∴A'F=CB'. A'C=B'F

∵CA'⊥CA，CB'⊥CB？ ∴∠a'cb'+∠acb=360— 180 = 180

And ∠ a' CB'+∠ ca 'f = 180 (the complementary angle of the inner angle on the same side),? ∴∠CA'F=∠ACB

By CB'=CB, CB'=A'F? , ? Get A'F=CB,

In △CA'F and △CAB, CA'=CA, A'F=CB,? ∠CA'F=∠ACB

∴△CA'F≌△CAB(SAS)，？ ∴CF=AB

And CE= 1/2CF,? ∴CE= 1/2AB。

(2) Is there a minimum value:

There is a minimum value. As shown in the figure, the size of OC is related to the size of ∠ACB of triangle ABC. When ∠C is an acute angle, the O point is in the shape.

When ∠C is obtuse, it is outside the △ABC shape. When △ABC is a right triangle, the connected A'B and AB' are on the same straight line with CA' and CB', so the intersection point O of A'B and AB' coincides with point C, and the distance between point O and point C is the smallest, so the minimum value is zero. (Independent of the given two data 5 and 8)