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.


Let's make following construction. $E$ is point of $BC$ such that $BE=BD$, $F$ is point symmetric to $E$ around $CD$. $G$ is point symmetric to $D$ around $AB$. Angular measures which can be restored from problem statement are given in picture.

$\angle DFC=100$°$=2\angle DAC$, then $F$ is center of circumcircle of triangle $ACD$. Then $FA=FD=FC=DE=CE$. Triangle $BDG$ is congruent to triangle $BDE$, then $DG=DE$.
$AF=FD$, $AD=AG$. Then $A$ is common point of perpendicular bisector of $DG$ and circle with center $F$ and radius $FD$.
Let point $H$ is symmetric to $D$ around $FG$, then $FDGH$ is rhombus. Because of $\angle FDG=120$° in this rhombus $DH=GH$ and $H$ is on perpendicular bisector of $DG$ which is $BA$. Then $\angle BHD=30$°, $\angle BHF=90°$ and $FH=FD$. Then perpendicular bisector of $DG$ is tangent to circle with center $F$ and radius $FD$. As tangent has only one common point with circle, then $A=H$ and $\angle BAD=\angle BHD=30$°.
