Try solving this with pure Euclid!

Consider point $E$ such that $\triangle$ $ADE$ is an equilateral triangle and $DE$ does not intersect with $AB$ . Then $\triangle$ $ABC$ $=$ $\triangle$ $BEA$ $(1)$ . From $(1)$ we get that $\triangle$ $ADB$ $=$ $\triangle$ $EDB$ $(2)$ . From $(1)$ and $(2)$ we get that $\angle ABD=10^{\circ}$ . Thus, $\angle BDC$ $=$ $\angle ABD$ $+$ $\angle BAC$ $=$ $10^{\circ}$ $+$ $20^{\circ}$ $=$ $30^{\circ}$ .

Draw an equilateral triangle inside ABC triangle. BAC=ACI=20, AB=AC, AD=BC=IC => ABD=CAI (s-a-s) => ABD=CAI=10 => BDC=10+20=30

In all Euclidean solutions that are already posted, an ${\color{#3D99F6}{\text{equilateral triangle}}}$ is always the key. So, here is another one:

Let $c$ be the circle with center $A$ and radius $AB=R$ . $AC=AB=R$ , hence, $C$ lies on $c$ . Consider point $E$ on $c$ , such that $\triangle {\color{#3D99F6}{ABE}}$ is equilateral and overlapping $\triangle ABC$ .

$\left. \begin{matrix} \angle BAC=20{}^\circ \\ \angle BAE=60{}^\circ \\ \end{matrix} \right\}\Rightarrow \angle CAE=40{}^\circ \Rightarrow \overset\frown{CE}=40{}^\circ \Rightarrow \angle CBE=20{}^\circ$

Since $BC=AD$ , $BE=AE$ and $\angle CBE=\angle BAD$ , $\triangle CBE=\triangle BAD$ (s-a-s), thus, $\theta=\angle ABD=\angle BEC \ \ \ \ \ (1)$

Alongside, $\angle BAC=20{}^\circ \Rightarrow \overset\frown{BC}=20{}^\circ \Rightarrow \angle BEC=10{}^\circ \overset{(1)}{\mathop{\Rightarrow }}\,\theta=10{}^\circ$ .

Finally, $\varphi =\angle BDC=\theta +\angle BAC=10{}^\circ +20{}^\circ =30{}^\circ$ .

The answer is $\boxed{30}$ .

link text

$WLOG~let~BC=2.~~AB=AC=\dfrac{BC}{cos80}\\ DC=AB-BC.~~DB~from~DC,~DB~and~included~angle~C~with~Cos~Law.\\ Use~Sin~Law~in~\Delta~DBC~calculate~\angle~CDB.\\ \therefore~\angle~CDB~=~asin~\left ( \dfrac{BC*sin80} {\sqrt{2^2+(\frac 2 {cos20})^2-2*2*\frac 2 {cos20}*cos(80)}} \right )=~30^o.\\ If ~Sin~and~Cos~Laws~are~a~part~of~Trig.,~I ~will~post~other~ solution~soon.$

Firstly, $\angle ABC = \angle ACB = 80°$

Now, add a point $E$ , such that $\triangle ABC \cong \triangle EAD$ and $\triangle EAD$ doesn't overlap $\triangle ABC$ (i.e. make a copy of $\triangle ABC$ and position it so that its base completely overlaps segment $AD$ , since $AD$ = $BC$ ). since $\angle ABC + \angle EAB = \angle ABC + \angle EAD + \angle BAC = 180°$ , we have that segments $BC$ and $AE$ are parallel. Now, add another point $F$ on line $BC$ such that $ABFE$ is a parallelogram. Not only will $ABFE$ be a parallelogram, but it will be a rhombus as well because $AB = AE$ (from congruency).

Now we have lots of segments of same length:

$AB = AC = EA = ED = EF = BF$

Because $ABFE$ is a rhombus: $\angle AEF = \angle ABF = 80°$ , so $\angle DEF = \angle AEF - \angle AED\ = 60°$ . With $DE = EF$ and $\angle DEF$ , we find that $\triangle DEF$ is equilateral, so we can add $DF$ to the equality above:

$AB = AC = EA = ED = EF = BF = DF$

Now, since $BF = DF$ , $\triangle BDF$ is isosceles with $\angle BFD = \angle BFE - \angle AFE = 100° - 60° = 40°$ and $\angle DBF = \angle BDF = 70°$ .

In the end, we easily find that $\angle BDC = 180° - \angle BCD - \angle DBC = 180° - \angle BCA - \angle DBF = 180° - 80° - 70° = \boxed{30°}$

Let $AB=AC=1$ . By sine law on $\triangle ABC$ ,

$\dfrac{BC}{\sin 20}=\dfrac{1}{\sin 80}$ $\implies$ $BC=\dfrac{\sin 20}{\sin 80}$

By cosine law on $\triangle ABD$ ,

$(BD)^2=\left(\dfrac{\sin 20}{\sin 80}\right)^2+1^2-2\left(\dfrac{\sin 20}{\sin 80}\right)(1)(\cos 20)$

$BD=\sqrt{\left(\dfrac{\sin 20}{\sin 80}\right)^2+1-\dfrac{2 \sin 20 \cos 20}{\sin 80}}$

By sine law on $\triangle BDC$ ,

$\dfrac{\sin \angle BDC}{BC}=\dfrac{\sin 80}{BD}$

$\dfrac{\sin \angle BDC}{\dfrac{\sin 20}{\sin 80}}=\dfrac{\sin 80}{\sqrt{\left(\dfrac{\sin 20}{\sin 80}\right)^2+1-\dfrac{2 \sin 20 \cos 20}{\sin 80}}}$

$\angle BDC = \boxed{30^\circ}$