O is the center of a circle. OBCD is a rhombus inscribed in the circle. If OB=BC=CD=DO, find the angle BAD.

O is the center of the circle, so all segments from O to points on the circumference are radii. Thus, $OA = OB = OC = OD$.
OBCD is a rhombus. By definition, all sides of a rhombus are equal in length. So, $OB = BC = CD = DO$.

Combining the properties from Step 1, we have:
$OB = OC = OD$ (radii)
$OB = BC = CD = DO$ (sides of rhombus)
Therefore, all segments $OB, BC, CD, DO, OC$ are equal in length.
This implies that $\triangle OBC$ is an equilateral triangle (since $OB=BC=OC$) and $\triangle OCD$ is an equilateral triangle (since $OC=CD=DO$).

In an equilateral triangle, all interior angles are $60^\circ$.
So, $\angle BOC = 60^\circ$ and $\angle COD = 60^\circ$.

The angle $\angle BOD$ is the sum of $\angle BOC$ and $\angle COD$.

This angle $\angle BOD$ is the angle at the center subtended by the arc BCD.

The angle subtended by an arc at the center of a circle is twice the angle subtended by the same arc at any point on the remaining part of the circumference.
In this case, $\angle BAD$ (which is $t$) is the angle at the circumference subtended by the arc BCD.
Therefore,

The value of $t$ is $\boxed{\text{60}^\circ}$.