1) It is necessary to make the circumscribed circle of △ABC intersect with CD at H' (since AD=BE, CD will not be tangent to the circle), and then prove that H' and H coincide.

2) This question is to prove that △AKH is an equilateral triangle, and △ABK and △ACH are congruent, which is great;

Make an auxiliary line, take a little g' on HC, make HG'= KG (this is the most critical point in this question), and connect AG', then △AGK and △AG'h are also congruent).

Look at △BEG and △DAG' (we want to prove that they are congruent). ∠BGE=∠AGK=∠AG'D, called BE=AD, we still need to find a diagonal line. Pay attention to ∠CBH and ∠D, ∠ D =180-∠ DHB-∠ DBH = 60-∠ DBH = ∠ CBH. This proves that everything is equal.

So BG=DG'= 1+2=3, CH=BK=3+ 1=4, so CD=6.

Hey, I'm old. It took me a long time to work out these contest questions.

I didn't say a ball! ! ! attached drawing