Two triangles are similar if the ratios of the corresponding sides of the two groups are equal and the corresponding included angles are equal.

Let the side length of the square in the graph be a, that is, AB=BC=CD=DE=a, which is obtained by Pythagorean theorem: AC= root number 2 * A.

△ACD and△ ACE, the corresponding angle ∠ACD=∠ACE,

The edge CD of △ACD corresponds to the edge AC in △ACE, and the edge AC in △ACD corresponds to the edge CE in △ACE.

Corresponding edge AC/CD= radical number 2*a/a= radical number 2, CE/AC=2a/ radical number 2*a= radical number 2,

So AC/CD=CE/AC, and ∠ACD=∠ACE, you get △ACD∽△ECA.

(2) If ∠ 1 = 45, ∠ ACD = 135 and ∠ 3+∠ CAE = 45 in square ABCH,

From △ACD∽△ECA, ∠2=∠CAE, ∠3=∠CAD,

Then ∠ 3+∠ CAE = ∠ 3+∠ 2 = 45, so ∠ 1+∠ 2+∠ 3 = 90.