How do I prove ABCD to be a rectangle

Answer:
steps below
Step-by-step explanation:
AB = DC, AE=DE, ∠BAE = ∠CDE = 90°
ΔBAE ≅ ΔCDE (SAS)
BE = CE .. corresponding sides of congruent triangles
∠EBC = ∠ECB ... base angles of isosceles triangle
∠ABC = ∠ABE + ∠EBC = ∠DCE + ∠ECB = ∠DCB
∠ABC + ∠DCB = 360° - ∠CDE - ∠BAE = 360° - 180° = 180°
∠ABC = ∠ DCB = 1/2 * 180° = 90°
∴ ABCD is rectangle