Need pure geometric solution for proof on 10-20-40-50 angle problem $D$ is a point in $\triangle{ABC}$ so that $\angle{ABD}=10^{\circ}$, $\angle{DBC}=20^{\circ}$, $\angle{BCD}=40^{\circ}$, $\angle{DAC}=50^{\circ}$.
Find $\angle{BAD}$.
This problem is easily done with trigonometric Ceva theorem as:
$$
\begin{aligned}
&\dfrac{\sin x}{\sin50^{\circ}}
\cdot
\dfrac{\sin(60^\circ-x)}{\sin40^{\circ}}
\cdot 
\dfrac{\sin20^\circ}{\sin10^\circ}=1 
\\
\implies 
&
\sin x \cdot \sin(60^\circ-x)
=
\dfrac
{\sin50^\circ\cdot \sin40^\circ\cdot \sin10^\circ}
{\sin20^\circ}
\\
&\qquad
=
\dfrac
{ \cos40^\circ\cdot \sin40^\circ \cdot \sin10^\circ}
{2\sin10^\circ\cdot \cos10^\circ}
=
\dfrac{\sin80^\circ}{4\cos10^\circ}
=\dfrac{1}{4}
\\
\implies
& -\dfrac{\cos60^\circ - \cos(2x-60^\circ)}2 = \dfrac{1}{4}
\\
\implies &
\cos(2x-60^\circ)=1 
\\
\implies 
&2x-60^\circ=0
\\
\implies 
&x=\boxed{30^\circ}\ .
\end{aligned}
$$
Any idea on how to solve this problem in pure geometric approach? Thanks.



I just got this solution (no phantom point and few verbal words):
Let $E$ be the circumcenter of $\triangle{ACD}$, so $AE=CE=DE, \angle{DEC}=2\angle{DAC}=100^{\circ}$
$\implies \angle{EDC}=\angle{ECD}=40^{\circ}=\angle{BCD} \implies DE \parallel BC$
Extend $ED$ and intersects with $AB$ at $F$, so $\angle{AFE}=\angle{ABC}=30^{\circ}$. Make $G$ on $BC$ so that $\angle{EFG}=100^{\circ}=\angle{FEC} \implies CEFG$ is an isosceles trapezoid $\implies FG=CE$
Let $H$ be the circumcenter of $\triangle{BFG}$, so $BH=GH=FH, \angle{FHG}=2\angle{FBG}=60^{\circ} \\ \implies \triangle{FGH} \text{ is an equilateral triangle}\\ \implies FH=GH=BH=FG=CE=DE=AE, \angle{HFG}=60^{\circ} \\ \implies \angle{HBF}=\angle{HFB}=\angle{HFG}-\angle{BFG}=\angle{HFG}-(180^{\circ}-\angle{FBG}-\angle{FGB}) \\ =\angle{HFG}-(180^{\circ}-\angle{FBG}-\angle{GFE}) =60^{\circ}-(180^{\circ}-100^{\circ}-30^{\circ})=10^{\circ} \\ \implies FH \parallel BD, \angle{HBD}=20^{\circ}=\angle{DBC}=\angle{FDB} \\ \implies BDFH \text{ is an isosceles trapezoid} \\ \implies BH=DF \implies DF=DE=AE$
Check $\triangle{AEF}, \angle{AFE}=30^{\circ}, AE=DE=DF \implies EF=2AE$.
Let $O$ be the circumcenter of $\triangle{AEF}$, so $\angle{AOE}=2 \angle{AFE}=60^{\circ}, AO=FO=EO$
$\implies \triangle{AOE} \text{ is equilateral triangle}\\ \implies AO=OE=AE=DE=DF=FO=EO \\ \implies AD=AO=FD \\ \implies \angle{BAD}=\angle{AFE}=\boxed{30^{\circ}}$
