In $\triangle A B C$, $\mathrm{D}$ is a point on $\mathrm{BC}$ such that $\angle B A D=\frac{1}{2} \angle A D C$ and $\angle B A C=77^{\circ}$ and $\angle C=45^{\circ}$. What is the measure of $\angle A D B$ ?

64°

In \(\Delta \)ABC,

\(\angle\)BAC + \(\angle\)ABC + \(\angle\)ACB = 180  [Angle sum property]

= 77 + \(\angle\)ABC + 45 = 180

= 122 + \(\angle\)ABC = 180

= \(\angle\)ABC = (180 - 122)

= \(\angle\)ABC = 58

In \(\Delta \)BAD,

= \(\angle\)ABD + \(\angle\)BAD + \(\angle\)ADB = 180 [Angle sum property]

= 58 + 58 + \(\angle\)ADB = 180

= 116 + \(\angle\)ADB = 180

= \(\angle\)ADB = 180 - 116

= \(\angle\)ADB = \({64}^\circ\)

Therefore, \(\angle\)ADB is \({64}^\circ\).