Connect OE, BE

∵∠ABC = 90°, AB ∵∠ O diameter.

∴BC intercepts point ⊙ O at point B, ∠ AEB = 90, ∠ BEC = 90, ∠ BAE+∠ Abe = 90.

∴∠EBC=∠BAE

∫∠BEC = 90°, and D is the midpoint of BC.

∴BD=DE=DC

∴∠EBD=∠BED

∴∠BED=∠EBD=∠BAE

In ∵△ABE, O is the midpoint of hypotenuse AB.

∴OE=OB=OA

∴∠OEB=∠OBE

∴∠OEB+∠BED=∠OBE+∠BAE=90

∴OE⊥DE

And o is the center of the circle, and ∴DE is tangent to ⊙ o.

Finish the certificate ~