In a circle with center $\mathrm{O}, \mathrm{AB}$ is a diameter and $\mathrm{CD}$ is a chord such that $\angle \mathrm{ABC}=34^{\circ}$ and $\mathrm{CD}=\mathrm{BD}$. What is the measure of $\angle \mathrm{DBC}$ ?

28°

\(\angle\)ACB = \({90}^\circ\)

In \(\Delta \)BCD

\(\angle\)B = \(\angle\)C = \(\angle\)DBC due to BD = CD

In cyclic quadrilateral ABDC

\(\angle\)ACD + \(\angle\)DBA = \({90}^\circ\)

= \({90}^\circ\) + \(\angle\)DBA + \({34}^\circ\) + \(\angle\)DBA = \({180}^\circ\)

= 2\(\angle\)DBA = 180 - 124

= \(\angle\)DBA = \(\frac{56}{2}\)

= \(\angle\)DBA = \({28}^\circ\)

Therefore, \(\angle\)DBA is \({28}^\circ\)