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.

 A: 
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 $\angle{AOE}=2 \angle{AFE}=60^{\circ}, AO=FO=EO$
$\implies \triangle{AOE} \text{ is equilateral triangle}\\ \implies AO=OE=AE=DE=DF=FO=EO \\ \implies AD=AO=FD \\ \implies \angle{BAD}=\angle{AFE}=\boxed{30^{\circ}}$
A: I'll give a first quick solution, one can stop after it if the problem implicitly gives the information that the configuration to be realized exists and is unique. Some discussion follows to cover also this point and  have a complete solution, that cannot lose points e.g. in the conditions of an olympiad.

It is natural to try to realize the given picture using vertices $A,B,C$ among those of a regular poligon. It turns out, we need a regular $18$-gon, in the picture the vertices have the labels from $\Bbb Z$ modulo $18$, and for easy using as a point there is a label $k'$ instead of $-k$ for $-k$ from $-8=10$ to $-1=17$. We place the letters $A,B,C$ at positions $0,7,3'$, and show that the point $D$ defined as $3'3\cap 71'$ (so that $\Delta DBC$ has angles of $20^\circ$, $40^\circ$ in $B,C$) lies on the chord from $A=0$ making the $50^\circ$ angle with $AC$, which is $08'$.

In other words, we have to show:
Proposition $(P)$: In the given regular $18$-gon the lines $08'$ and $71'$ and $3'3$ intersect in a point $D$.
Proof: Let $I$ be the incenter of $\Delta 173'=\Delta 1BC$. So $I=17'\cap 71'\cap 3'4$, and since $11'7'7$ is an isosceles trapezium, it is also its symmetry axis $09$. Now observe that $D$ is the orthocenter of $\Delta IAC=\Delta I03'$, the heights are the three lines to be shown concurrent:
$$
D=
\underbrace{\ 7I1'\ }_{\perp\text{ on } 03'}\cap
\underbrace{\ 3'3\ }_{\perp\text{ on } 9I0}\cap
\underbrace{\ 08\ }_{\perp\text{ on } 3'I4}\ .
$$
$\square$
The angle $\widehat {BAD}$ "sees an arc covering three sides", so it is $\color{blue}{\boxed{30^\circ}}$.


We are done, but
let's turn the above argument into a constructive solution not assuming that the given constellation exists and is unique.
We make the Proposition $(P)$ work in this context.
So we know by it that $08'\cap 71'\cap 3'3$ is a point, call it $D$.
We restart with what we certainly have, this is
the triangle $\Delta BCD$, it is fully known (modulo similarity), we are drawing it, so that $B,C$ are placed in $7,3'$.
Based on it, we have to construct a point $A$ such that $\widehat{DAC}=50^\circ$,
a point located in the half-plane w.r.t. $CD$ not containing  $B$.
The locus of such a point is an arc of a circle, call this circle $\omega$, its center $\Omega$. We know that the vertex $0$ is on this circle, so $\omega=(D0C)$. I need the picture (without any point $A$ on it):

Now consider a moving ray, in the radar with origin in the point $B$.
It intersects $\omega$ in either two points, or in one point, or in no point. We fix this ray to make an angle of $10^\circ$ with $BD$, as given.
This ray intersects $(\omega)$ in at least one point, $0$, as guaranteed by $(P)$. Does $B0=70$ intersect $\omega$ also in an other point, or it is tangent to it?! We show:
Lemma: In the above situation, $B0$ is tangent to $\omega$ in $0$.
(So $A=0$ is the unique solution to our problem.)
Proof of the Lemma: $\Delta D0C$ has angles of $30^\circ$, $50^\circ$ in $C$, resp. $0$. So $\widehat {0\Omega C}=2(50^\circ+30^\circ)=160^\circ$, which results in angles of $10^\circ$ in $0,C$ in $\Delta 0\Omega C$. This gives:
$$
\widehat{B0\Omega}=
\widehat{B0D}+
\widehat{D0C}+
\widehat{C0\Omega}
=
30^\circ +
50^\circ +
10^\circ 
=
90^\circ 
\ .
$$
$\square$
A: 
We use this fact that the mirrors of circumcircle about the sides of triangle meet at it's orthocenter.
Step 1:Draw circumcircle(1) of ABC center is O.Draw altitude AK.Mirror AB, AK, circle (1) about base BC, circle (2) is mirror of circle (1). Altitude AK  intersects the circle 2  at E. E is orthicenter of triangle ABC. Connect D to E and extend it to touch AC at H. We show that DH is collinear with BD and in this way BH is altitude of triangle,Clearly $\angle BA'A=60^o$
In circle (2):
$\overset{\large\frown}{BE}=120^o$
$\overset{\large\frown}{EC}=40^o$
$\Rightarrow \overset{\large\frown}{BA'C}=360-(120+40)=200^o$
$\overset{\large\frown}{A'C}=60^o$
$\Rightarrow \overset{\large\frown}{BA'}=140^o$
That is if we connect B to E we have:
$\angle BEA'=\frac {140}2=70^o$
This ls competent with situation in right angled triangle BEK where $\angle BEK =180-(90+20)=70^o$. This means that BD and BE are collinear and  BH is an altitude of triangle.
Step 2:Draw circumcircle (3) of triangle BCH, BC will be it's diameter. Extend CD to meet the circle (3) at point F.We have:
$\overset{\large\frown}{CH}=40^o$
$\overset{\large\frown}{BF}=80^o$
$\Rightarrow$$\overset{\large\frown}{FH}=60^o$
$\Rightarrow \angle FCH=30^o$
which finally gives $\angle BAD=30^o$
Update:  All triangles has this property that the mirror of their circumcircle about it's sides meet at  orthocenters of the them. See following figure:

A: 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 $\angle BAD=\angle BHD=30$°.
