That unknown angle

Figure 1 We first consider a regular 18-gon constructed on side $AB=s$ of $\triangle ABC$ , having $A$ as its circumcenter. Its vertices are labelled as seen on figure 1. We notice that diagonals ${{A}_{5}}{{A}_{16}}$ and ${{A}_{6}}{{A}_{15}}$ are parallel. Moreover, diagonal $C{{A}_{11}}$ is parallel to side ${{A}_{15}}{{A}_{16}}$ . Hence $AE{{A}_{16}}{{A}_{15}}$ is a parallelogram.

Thus, $AE={{A}_{15}}{{A}_{16}}\ \ \ \ \ (1)$ Figure 2

In figure 2, ${{A}_{5}}{{A}_{16}}\parallel {{A}_{6}}{{A}_{15}}$ and ${{A}_{12}}{{A}_{16}}\parallel {{A}_{9}}B$ , hence $GH{{A}_{12}}F$ is a parallelogram, thus, $GF=H{{A}_{16}}$ .

Furthermore, due to pairs of parallel lines, $\angle GFA=\angle H{{A}_{16}}H$ and $\angle FGA=\angle{{A}_{16}}H{{A}_{15}}$ . Consequently, by ASA, $\triangle FGA=\triangle {{A}_{16}}H{{A}_{15}}$ ,
which gives that $AF={{A}_{15}}{{A}_{16}}=s\ \ \ \ \ (2)$ Figure 3

$(1)$ and $(2)$ imply that both diagonals $C{{A}_{11}}$ and $B{{A}_{9}}$ intersect $AC$ at the given point $D$ (figure 3). Because of this, $m\angle ABD=\dfrac{m\overset\frown{{{A}_{9}}{{A}_{10}}}}{2}\Rightarrow \angle ABD=\boxed{10{}^\circ}$ .

Let x = $\overline{BD}$ , and $y = \overline{AD} = \overline{BC}$ and $\theta = \angle ABD$ .

Noting that $\angle DCB = 80^{\circ}$ , then by applying the Law of Sines to $\triangle ABD$ and $\triangle DBC$ , we get,

$\dfrac{\sin 20^{\circ}} { x} = \dfrac{ \sin \theta} { y }$

and

$\dfrac{\sin 80^{\circ}} { x} =\dfrac{ \sin (20^{\circ} + \theta) }{ y}$

Dividing the two equations out, yields,

$\dfrac{\sin 20^{\circ}}{ \sin 80^{\circ} } = \dfrac{ \sin \theta }{ \sin (20^{\circ} + \theta) }$

Cross multiplying,

$\sin 20^{\circ} \sin (20^{\circ} + \theta) = \sin 80^{\circ} \sin \theta$

Using the identity $\sin(A + B) = \sin A \cos B + \cos A \sin B$ , the above equation becomes,

$\sin 20^{\circ} ( \sin 20^{\circ} \cos \theta + \cos 20^{\circ} \sin \theta ) = \sin 80^{\circ} \sin \theta$

Dividing through by $\cos \theta$ , and collecting terms,

$\tan \theta = \dfrac{ \sin^2 20^{\circ} } { \sin 80^{\circ} - \sin 20^{\circ} \cos 20^{\circ} } = 0.1763269807084$

Therefore, $\theta = \tan^{-1} 0.1763269807084 = \boxed{10^{\circ} }$

Rotate side $AB$ $40^{\circ}$ clockwise around point $A$ such that $B$ ends up in a new position $B'$ . Draw $B'C$ and $B'B$ .

Observe that, in $\bigtriangleup AB'C$ , $AB'=AB=AC$ and $\angle B'AC$ $=$ $40+20=60^{\circ}$ , which indicates that $\bigtriangleup AB'C$ is equilateral.

Also, observe that, in $\bigtriangleup B'BC$ & $\bigtriangleup DBA$ , $BC=AD$ (given), $B'C=AB'=AB$ & $\angle B'CB$ $=$ $\angle ACB$ $-$ $\angle ACB'$ $=$ $80-60=20^{\circ}$ $=$ $\angle BAD$ .

$\Rightarrow$ $\bigtriangleup B'BC$ $\cong$ $\bigtriangleup BDA$ (By $A-S-A$ criterion of congruence)

Hence, $\angle ABD$ $=$ $\angle CB'B$ $=$ $\angle AB'B$ $-$ $\angle AB'C$ $=$ $\frac{180-40}{2}$ $-$ $60$ $=$ $70-60$ $=$ $\boxed {10^{\circ}}$ .