we know that
the triangle BDC it's an isosceles triangle
m ∠BDC= m ∠DBC-----> 45°
so
m∠DCB=180-(45+45)-----> 90°
we know that
The measure of the external angle is the semi difference of the arcs that it covers.
the arcs that it covers are
arc BD=90 (central angle 90°)
arc 360-90=270°
so
m∠DAB=(1/2)*[270-90]----> 90°
the answer is
m∠DAB=90°
the figure ABCD is a square
