In the given figure, AB is the diameter of the circle with center O. If \ BCD = 110^ and chord CD = chord CB, find the value of t (i.e. \ ADC).

The figure shows a cyclic quadrilateral ABCD inscribed in a circle with center O.

• AB is a diameter because it passes through the center O.
• $\angle BCD = 110^\circ$.
• $\angle ADC = t$.
• Chord CD = Chord CB (indicated by the double hash marks).

Since AB is a diameter, the angle subtended by the diameter at any point on the circumference is $90^\circ$.
Therefore, $\angle ACB = 90^\circ$.

We know $\angle BCD = \angle BCA + \angle ACD$.
Substitute the known values:

Given that chord CD = chord CB.
Equal chords subtend equal arcs. Therefore, arc CD = arc CB.
Angles subtended by equal arcs at the circumference are equal.
So, $\angle CAD = \angle CAB$.

In a cyclic quadrilateral, opposite angles sum to $180^\circ$.

Substitute the given value for $\angle BCD$:

We know $\angle DAB = \angle CAD + \angle CAB$.
Since $\angle CAD = \angle CAB$ (from Step 4), we can write:

The sum of angles in a triangle is $180^\circ$.
In $\triangle ADC$:

Substitute the values for $\angle CAD$ and $\angle ACD$:

The value of $t$ is $125^\circ$. This corresponds to option D.

The final answer is $\text{D. } 125^\circ$.