(1) In Figure 7, the bisector AE of ∠BAD is drawn with a ruler (drawing traces are left, but no writing method is used), which proves that the quadrangle ABED is a diamond;
(2)ABC = 60, EC=2BE, verification: ED⊥DC.
(1) Solution: Take point B and point D as the center respectively, make an arc with the length greater than AB as the radius, the two arcs intersect at point P, and then connect AP, that is, the bisector with AP ∠BAD, and AP and BC intersect at point E,
∵AB=AD,∴△ABO≌△AOD ∴BO=OD
∫ AD//BC, ∴∠OBE=∠ODA, ∠OAD=OEB.
∴△BOE≌△DOA
∴BE=AD (parallel and equal)
∴ quadrilateral ABDE is a parallelogram, AB=AD,
A quadrilateral is a diamond.
(2) Let DE=2a, then CE=4a, and the intersection D is DF⊥BC.
∵∠ ABC = 60, ∴∠ DEF = 60, ∴∠ EDF= 30, ∴ EF = DE = A, then DF=, CF=CE-EF=4a-a=3a,
∴
∴DE=2a, EC=4a, CD=, forming a set of Pythagorean numbers,
∴△EDC is a right triangle, and then ED⊥DC.