=BD+CD+CB, so the perimeter of the triangle BCD is also 4~

2. Make the bisector OT of the angle AOB, connect MN, make the middle vertical line XY of MN, and the intersection of XY and OT is P, which is the demand ~

3. Let AB and MN intersect at point E, and CD and MN intersect at point F ... Connect ME and ne, and it is easy to prove that all triangle EMFs are equal to triangle EMN, then: ME=NE, ∠MEF=∠NEF, ∠ AEM+∠ MEF = 90 = ∠ NEF+∠.

If you don't understand the answer, ask again. Personally, I think I have made it very clear ~ give points ~