As shown in the figure, O is the center of square ABCD, BE bisects ∠DBC, passes through DC at point E, extends BC to point F, makes CF=CE, connects DF and passes through the extension line of BE.

EH⊥BD passes through point E at point H.

Step one:

It is easy to get CE = eh = HD, BH = BC, BH+DH = √ 2bc.

Step two:

It is proved that △ BCE △ DCF obtains Be = DF.

Step 3:

It is proved that BG bisects DF vertically. If DF = 2dg, then BE = DF = 2bg.

Step 4:

Using Pythagorean Theorem BH? +Eh? = Yes? Find the side length BC

Finally, find the square area.