Oops! Someone Gave Out A Clue

Let's make up a proof.

PROOF 1 (My approach)

FIRST , $\angle{BOC}$ is a supplement of $\angle{BOE}$ . We know that supplementary angles have their measurements total up to $180^{\circ}$ . Furthermore, alone, $\angle{EOC}$ measures $180^{\circ}$ . Therefore, $m\angle{BOC}+m\angle{BOE}=180^{\circ}$ . Since $m\angle{BOC}=105^{\circ}$ , then we should subtract it from $180^{\circ}$ : $180^{\circ}-105^{\circ}=75^{\circ}$ $\Rightarrow m\angle{BOE}$

SECOND , line $AD$ $\parallel$ line $BO$ . Since line $AD$ $\parallel$ line $BO$ , then we can also conclude that $m\angle{AFE}=m\angle{BOE}=75^{\circ}$ and $m\angle{AFC}=m\angle{BOC}=105^{\circ}$ , as shown in the images below:

THIRD , extend line $AD$ beyond the circle. After that, draw a tangent line that is parallel to line $EC$ . We call this tangent line as line $XY$ .

Notice that the measurements depicted in the section of line $EC$ equals the measurements formed on the section of line $XY$ . Then, by the application of the fact that "opposite angles have the same measurements," we can now find the $m\angle{ADB}$ . .

Hence, the answer is $\boxed{30^{\circ}}$ .

PROOF 2

Another way to settle up a proof is by using the exterior angle property of triangles .

$\angle{EOB}=180^{\circ}-105^{\circ}=75^{\circ}$ (linear pair).

Then, $\angle{DBO}=105^{\circ}-75^{\circ}=\boxed{30^{\circ}}$ (By exterior property of triangles).

*Note: $\angle{ADB}$ and $\angle{DBO}$ are co-interior angles.

In fact, supposing that AD // BO, <ADB can be anything between 0° (A on B) and 37.5° (A on E).