This constitutes an isosceles right triangle.

So ab 2 = AE 2+EB 2.

AE=AC+DB=3

So AB=3 times the root number 2.

(2) Make the middle vertical line of AB first,

Cross AB at o point, cross CD at p point,

Connect PA and PB, then PA=PB, which is the point to find.

The triangle AEB is an isosceles right triangle, and OP bisects AB vertically.

Then there is the three-point * * * line of E, O and P.

Angle CEP = 45, ECP = 90°.

So the triangle CEP is an isosceles right triangle with CP=CE=DB=2.

So DP=CD-CP=3-2= 1.