Let $ABC$ be an equilateral triangle.
$D$ is on $BC$ and $BAD$ angle is $20^{\circ}$.
Also let $I_1$, $I_2$ be the inner centers of triangle $ABD$, $ACD$ respectively.
$E$ is a point making the triangle $I_1I_2B$ be equilateral ($D$ and $E$ are on the opposite side of the line $I_1I_2$.
Then what is the angle $ADE$?
This is my question and I tried to find $4$ cyclic points or something like that, but I have no idea.
Please help me.
This is a late answer to this nice geometry problem... Here is a picture of the given geometric situation, together with auxiliary points (which will be introduce one by one in the sequel):
The problems gives the value $20^\circ$ for the angle $$ x=\widehat{BAD}\ . $$ It is convenient to introduce also the notation $$ y=\widehat{DAC}\ , $$ so that $x+y=60^\circ$. Let us work in a slightly more general case, and let $x$ be any angle (between $0^\circ$ and $30^\circ$). We will show the equality instead of $\overset?=$ below: $$ 30^\circ-x\overset?=\widehat{ADE}\ . $$ So for the given angle, $x=20^\circ$ we expect the answer $10^\circ=30^\circ-20^\circ=\widehat{ADE}$.
Here is one more picture for the case $x=14^\circ$, which avoids some coincidences compared to the above $20^\circ$ case:
The quadrilateral $ACDI_1$ is cyclic, because its angles in $D$, $80^\circ+50^\circ$, and $A$, $40^\circ+10^\circ$, add to $180^\circ$. In the general case we add $(120^\circ -y)+\frac 12(120^\circ -x)$ and $\frac 12x+y$. Denote the corresponding circle by $(1)$.
The quadrilateral $ABDI_2$ is cyclic, because its angles in $D$, $120^\circ+40^\circ$, and $A$, $20^\circ+20^\circ$, add to $180^\circ$. In the general case we have also the above computation, with $x,y$ exchanged. Denote the corresponding circle by $(2)$.
Let $J_1$ be the intersection $DI_2\cap BI_1$. Then: $$ \begin{aligned} \widehat{I_1J_1D} &= \widehat{BJ_1D} = 180^\circ- \widehat{J_1BD} - \widehat{J_1DB} \\ &= 180^\circ- 30^\circ - (120^\circ-x) -\frac 12(120^\circ-y) = \frac 12x= \widehat{I_1AD} \ , \end{aligned} $$ so $J_1$ is also on the circle $(1)=(AI_1D)$.
And $ \widehat{I_1DJ_1} =\widehat{I_1DI_2} = 90^\circ $ shows that the segment $I_1J_1$ is a diameter of $(1)$.
Let $J_2$ be the intersection $DI_1\cap CI_2$. Then: $$ \widehat{I_2J_2D} = \widehat{I_2AD} \ , $$ same argument (with $x,y$ exchanged), so $J_2$ is also on the circle $(2)=(AI_2D)$. And since $ \widehat{I_2DJ_2} =\widehat{I_2DI_1} = 90^\circ $ the segment $I_2J_2$ is a diameter of $(2)$.
Let $L_1,L_2$ be the centers of the circle $(1)$, and respectively $2$. Then $$\widehat{AL_1D} =2\widehat{ABD}=120^\circ=2\widehat{AI_1D}\ ,$$ so $L_1$ is on the circle $(1)$. Similarly: $$\widehat{AL_2D} =2\widehat{ACD}=120^\circ=2\widehat{AI_2D}\ ,$$ so $L_2$ is on the circle $(2)$.
Also, taking a look at the quadrilateral $AL_1DL_2$, it has perpendicular diagonals, the two triangles $\Delta AL_1D$, $\Delta AL_2D$ "building it" are isosceles, same basis, same opposed angle, $120^\circ$, so it is a rhombus. So the circles $(1)$ and $(2)$ are equal (congruent), same radius, and $\Delta AL_1L_2$, $\Delta DL_1L_2$ are equilateral.
Let us show now $J_1L_1\perp AI_2$. First of all, $J_1L_1$ is the angle bisector of $\widehat{AJ_1I_2}=\widehat{AJ_1D}$ since $L_1$ is the mid point of the $120^\circ$-arc $\overset \frown{AD}$. So $\widehat{AJ_1L_1}=30^\circ$. We also take a look at the $A$-angle $\widehat{I_2AJ_1}$. Cut it in two angles using $AC$. One angle is $y/2$. The other corresponds to the arc $\overset \frown{J_1C}$, so it is $\widehat{J_1DC}=(120-y)/2$. Putting all together, the needed $A$-angle is $60^\circ$. This gives the wanted perpendicularity. And moreover, $J_1L_1$ is the side bisector of $AI_2$.
In a similar fashion, $J_2L_2\perp AI_1$, and moreover $J_2L_2$ is the side bisector of $AI_1$. Consider now the circumcenter of $\Delta AI_1I_2$. It is $J_1L_1\cap J_2L_2$, the intersection of the side bisectors of the two sides, and the angle built with $I_1,I_2$ is twice $\widehat{I_1AI_2}=\frac 12x+\frac12y=\frac 12(x+y)=30^\circ$. So this circumcenter is the main actor from the exercise, it is $E$, making $\Delta EI_1I_2$ equilateral: $$ E=J_1L_1\cap J_2L_2\ , \qquad\text{ and } \ EA=EI_1=EI_2=I_1I_2\ . $$
Let $F$ be now the point of intersection of the lines $J_2EL_2$ and
$AD$.
Then
$$
\begin{aligned}
2\widehat{EFD}
&=
\overset{\frown}{AL_2} +
\underbrace{
\overset{\frown}{DBJ_2} +
\overset{\frown}{DI_2} }_{180^\circ} -
\overset{\frown}{DI_2}
=
2\widehat{ABL_2}+180^\circ-
2\widehat{DAI_2}
=
60^\circ + 180^\circ - y\ ,\text{ so}
\\
\widehat{EFD} &= 120^\circ -\frac y2\ .
\\[2mm]
\widehat{EL_1D} &=
\widehat{EL_1I_2} + \widehat{I_2L_1D}
=
\widehat{J_1L_1C} + 2\widehat{I_2J_2D}
=
\widehat{J_1DC} + 2\widehat{I_2AD}
\\
&=\frac 12(120^\circ -y)
+ 2\cdot\frac y2=60^\circ +\frac y2\ .\qquad\text{ This gives:}
\\[2mm]
180^\circ &= \widehat{EFD} + \widehat{EL_1D} \ .
\end{aligned}
$$
So the quadrilateral $EFDL_1$ is cyclic.
In a similar manner to the construction of $F=J_2EL_2\cap AD$, consider the point $G=J_1EL_1\cap AD$. The same argument gives $EGDL_2$ is cyclic, and the measure of the angle bigger $90^\circ$ built in $G$ by the lines $AD$ and $J_1EL_1$ is $120^\circ -\frac x2$.
So we know the angles in $\Delta EFG$. Things are depending on the position of $D$. We assume as in the pictures $x\le y$. (Else switch $B$ and $C$ to get them.) Then $\widehat{EGF}=\widehat{EGD}$ is the supplement of $120^\circ -\frac x2$, so it is $60^\circ+\frac x2$. And $\widehat{EFG}$ is the supplement of $120^\circ -\frac y2$, so it is $60^\circ+\frac y2$. This gives: $$ \begin{aligned} \widehat{FEG} &=180^\circ-\left(60^\circ +\frac x2+ 60^\circ +\frac y2\right)=60^\circ -\frac 12(x+y)=30^\circ\ ,\\ \widehat{L_1EL_2} &=\widehat{L_1EF} =180^\circ- \widehat{FEG} =150^\circ\ .\text{ Compare this information with:}\\ \widehat{L_1DL_2} &=60^\circ\text{ and $\Delta L_1DL_2$ equilateral.} \end{aligned} $$ This implies that $E$ is a point on the circle centered in $D$ passing through $L_1,L_2$. So: $$ \begin{aligned} \widehat{L_1DE} &= 2\widehat{L_1L_2E} = 2\widehat{L_1L_2A} - 2\widehat{EL_2A} = 120^\circ - 2\widehat{J_2L_2A} \\ &=120^\circ - 2\widehat{J_2DA}=120^\circ - \widehat{BDA} = \widehat{BAD}=x\ . \\ &\qquad\qquad\text{ This gives:} \\ \widehat{EDA} &= \widehat{L_1DA} - \widehat{L_1DE} = 30^\circ -x\ . \end{aligned} $$ We have computed the required angle.
$$\square$$
Knowing this, we immediately get $$ ED\perp\ BC\ . $$
Putting all together, and collecting some bonus properties, that follow from what we know, we can state as a bonus:
Proposition: Let $\Delta ABC$ be equilateral. Let $D$ be a point on $BC$. Consider $I_1,I_2$ the incenters of $\Delta ABD$, and respectively $\Delta ACD$. Denote by $x,y$ the measure of the angles $\widehat{BAD}$, $\widehat{DAC}$, $x+y=60^\circ$.
Then:
Let further $E,F,G$ be the points $E=L_1J_1\cap L_2J_2$, $F=AD\cap L_2J_2$, $G=AD\cap L_1J_1$.
Then:
Here is a final picture for the above situation:
We have the following bonus proof for the location of $E'$. Consider the (outer) Fermat-Torricelli point for the triangle $\Delta AI_1I_2$. The three equilateral triangles to be constructed on the sides are $\Delta AI_2J_1$, $\Delta AI_1J_2$, $\Delta I_1I_2E'$, so the three lines are concurrent: $AE'$, $I_1J_1=BI_1J_1$, $I_2J_2=CI_2J_2$. The last two lines are the medians / heights / bisectors from $B$, $C$, so they intersect in the center $O$ of $\Delta ABC$, so $A,O,E'$ are colinear.
$\square$
