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 $\angle \mathrm{DBC}$ ?
Answer & explanation
Correct answer: option 2
\(\angle\)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 \({28}^\circ\)