In a circle with centre O, AC and BD are two chords. AC and BD meet at E, when produced. If AB is a diameter and $\angle AEB = 36^{\circ}$. then the measure of $\angle DOC$ is:

108°

As AB is the diameter of the circle

⇒ \(\angle\)ABC = \({90}^\circ\)

Now, \(\angle\)ABC is the exterior angle of \(\Delta \)BCE

⇒ \(\angle\)ACB = \(\angle\)BEC + \(\angle\)CBE

⇒ \({90}^\circ\) = \({36}^\circ\) + \(\angle\)CBE

⇒ \(\angle\)CBE = \({54}^\circ\)

Now, we know that

The angle made by an arc at the center is double of the angle amde by the same arc at another point on the circumference in the major segment.

So, \(\angle\)COD = 2 x \(\angle\)CBE

⇒ \(\angle\)COD = 2 x \({54}^\circ\) = \({108}^\circ\)

Therefore, \(\angle\)COD is \({108}^\circ\).