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.