Prove that A₁D, B₁E, C₁F intersect at the same point Let ABC be a triangle.AA₁, BB₁, CC₁ are the angle bisectors of the triangle. ω is circumcircle of ABC. ω∩AA₁=A₂, ω∩BB₁=B₂, ω∩CC₁=C₂. Circumcircles of AB₁B₂, BC₁C₂, CA₁A₂ intersect with AB,BC and CA at D,E,F respectively. Prove that A₁D, B₁E, C₁F intersect at the same point.
My try:
After some angle chasing I have found out that C₂E=BC₂, A₂C=A₂F, B₂A=B₂D. And after considering 3 equal triangles. I found out that AB=AF, BD=BC, CA=CE. But after that I haven't managed to get anything. I am suspecting we might use Pascal's theorem since there are 6 points on circle ω. 
 A: First of all, let us recall on the following picture what we have:

We have:
$$
\begin{aligned}
\widehat{DAB_2}
&=
180^\circ - \widehat{BAB_2}
= \widehat{BCB_2}
\\
&
=\widehat{BCA} + \widehat{ACB_2}  
=\widehat{BCA} + \widehat{ABB_2}  
=\widehat{BCA} + \widehat{B_2BC}  
\\
&=\widehat{BCB_1} + \widehat{B_1BC}  
=\widehat{AB_1B}
\\
&=\widehat{ADB_2}\ .
\end{aligned}
$$
(At the last step we used $AB_1B_2D$ cyclic.)
Similarly, the other claimed equality of angles holds, so that the triangles $\Delta ADB_2$, $\Delta BEC_2$, and $\Delta CFA_2$ are isosceles. 
Now from $BA_2=A_2C=A_2F$ we obtain the equality of the triangles $\Delta ABA_2$ and $\Delta AFA_2$, showing the claimed $AB=AF$ from the OP.
It remains to show the concurrence, and to start the answer.

For me, the simplest solution now is to use barycentric coordinates. A short introduction to barycentric coordinates is:
Max Schindler, Evan Chen, bary-short.pdf
(This is distroying the geometry, but it is the simplest solution.) We compute immediately the needed barycentric coordinates, and the equations of $A_1D$, $B_1E$, $C_1F$:
$$
\begin{aligned}
A_1 &=(0:b:c)\ ,\\
B_1 &=(a:0:c)\ ,\\
C_1 &=(a:b:0)\ ,\\[2mm]
D &=(a:c-a:0)\ ,\\
E &=(0:b:a-b)\ ,\\
F &=(b-c:0:c)\ ,\\[2mm]
& A_1D\ :& c(a-c) x +acy -abz &= 0\ ,\\
& B_1E\ :& -bc x +a(b-a)y +baz &= 0\ ,\\
& C_1F\ :& cb x -cay +b(a-b)z &= 0\ .
\end{aligned}
$$
I will say some words about this. Skip please, if already in a good shape using barycentric coordinates, and go straightforward to the determinant at the end. 
A point $P$ has absolute barycentric coordinates $(x,y,z)$ w.r.t. the triangle $\Delta ABC$ with sides $a,b,c$ iff we can write
$$P = xA+yB+zC\ ,\qquad x+y+z=1\ . $$
This has a formal sense, as written. To have a quick sense, either identify $A,B,C$ with their affixes in the complex plane and use operations from $\Bbb C$, or considered it "vectorially" with a missing (tacitly chosen) origin $O$, then fill in to the equality $OP=x\cdot OA+y\cdot OB+z\cdot OC$. (Vectorial computation, $OP$ is here the vector $OP$, not its length.) 
Sometimes, $(x,y,z)$ is a specific expression with large denominator. It is simpler to ignore the denominator, so something like $(x:y:z)$ denotes $\left(\frac x{x+y+z},\frac y{x+y+z},\frac z{x+y+z}\right)$. (And $x+y+z\ne 0$.)
Now we compute the points above. I will do it formally, since i have to type. (Use complex numbers interpretation to have a sense of what follows.) 
The angle bisector theorem gives $|A_1B|:|A_1C|=c:b$. We rewrite successively $b|A_1B|=c|A_1C|$, $b(B-A_1) = -c(C-A_1)$, $bB+cC=(b+c)A_1$, $A_1=\left(a,\frac b{b+c},\frac c{b+c}\right)=(0:b:c)$. 
Corresponding formulas hold for $B_1$, $C_1$.
Let us compute also the barycentric coordinates for $D$. We start with $|BA|:|BD|=c:a$, and similarly we get $a|BA|=c|BD|$, $a(A-B)=c(D-B)$, $aA+(c-a)B=cD$, $D=(a:c-a:0)$.
The equation for the line $A_1D$ is obtained by taking the vector product of (the vectors built out of the coordinates of) $A_1$, $D$. Or we just verify the claimed equation with $A_1$ and $D$. 
The concurrence of $A_1D$, $B_1E$, $C_1F$ is now equivalent to the vanishing of the following determinant, Lemma 18 in loc. cit.:
$$
\begin{vmatrix}
c(a-c) & ac & -ab\\
-bc & a(b-a) & ab\\
bc & -ca & b(c-b)
\end{vmatrix}
\overset{(!)}{=\!=}
0
\ .
$$
This is an easy computation. In fact, we can also get the coordinates of the intersection point $X$,
$$
X=(ab:bc:ca)=\left(\frac 1c:\frac 1b:\frac 1a\right)\ ,
$$
and there is some symmetry in the asymmetry of this formula. (Its shape shows that $X$ is a "complicated point".)
$\square$
(A solution using Ceva / Menalaus can also be given.)

Later edit: The Ceva / Menelaus solution is based on the above knowledge of the point $X$. We construct $A_3$ in the following picture by intersecting the parallel from $C_1$ to $AC$ with $BC$:

Similarly we construct $B_3$, and $C_3$. Then all six lines $AA_3$, $A_1D$; $BB_3$, $B_1E$; $CC_3$, $C_1F$ are concurrent in $X$. I will drop maybe an other solution based on this observation.
A: Denote by $k$ the inscribed circle of triangles $ABC$ and let $I$ be its center. Let's focus on the quad $BEC_1C_2$.  Then $$\angle\, C_1AC =  \angle \, BAC = \alpha = \angle\, BC_2C = \angle \, BC_2C_1$$ Since $BEC_1C_2$ is cyclic, 
$$\angle \, C_1EC = \angle \, BC_2C_1 = \angle\, C_1AC = \alpha$$
However, $\angle \, ACC_1 = \angle \, BCC_1 = \angle \, ECC_1 = \frac{1}{2} \angle \, ACB$ which implies that triangles $ACC_1$ and $ECC_1$ are congruent and in fact mirror-symmetric with respect to the angle bisector $CC_1$. Since the incenter $I$ lies on the angle bisector $CC_1$, the mirror-symmetry with respect to $CC_1$ transforms the incircle $k$ to itself. However, line $AC_1 \equiv AB$ is tangent to $k$, so its image which is the line $EC_1$, is also tangent to the incircle $k$.   
We can apply the same arguments to the other two angle bisectors $AA_1$ and $BB_1$  and the respective quads $CFA_1A_2$ and $ADB_1B_2$,  and conclude that the pair of triangles $BAA_1$ and $FAA_1$ are mirror-symmetric with respect to angle bisector $AA_1$, and that   the pair of triangles $CBB_1$ and $DBB_1$ are mirror-symmetric with respect to angle bisector $BB_1$. Consequently, the pair of mirror-symmetric lines $FA_1$ and $BA_1$ are tangent to the incircle $k$, as well as the pair of mirror-symmetric lines $DB_1$ and $CB_1$ are also tangent to the incircle $k$.
Therefore the hexagon $EA_1FB_1DC_1$ is superscribed around the incircle $k$ of the triangle $ABC$. By Brianchon's theorem the diagonals
$$ A_1D, \,\, B_1E, \, \, C_1F$$
 of the hexagon $EA_1FB_1DC_1$ must intersect in a common point.
A: Here is an other answer, based on my already given answer using barycentric coordinates to get the location of the intersection point. Well, we already have an accepted answer, so i will make it short.
Let $a,b,c$ be the sides of the given triangle.
From the OP we know that $D$ is placed on $BA$ so that $\Delta BCD$ is isosceles, i.e. $BD=BC=a$. We construct $A_3$ on $BC$ so that $C_1A_3\| AC$. And similarly $B_3$, $C_3$. Let $U=BB_3\cap CD$. Picture so far:
 
As constructed, $AA_3$, $BB_3$, $CC_3$ are concurrent in a point $X$, reciprocal of the theorem of Ceva:
$$
\frac{A_3B}{A_3C}\cdot
\frac{B_3C}{B_3A}\cdot
\frac{C_3A}{C_3B}
=
-
\frac ab\cdot 
\frac bc\cdot 
\frac ca\cdot 
=-1\ .
$$
Let us show that $DA_1$ is also passing through $X$.

Menelaus for $\Delta ADC$ w.r.t. the transverse line $BB_3U$ gives
$$
1 = 
\frac{BD}{BA}\cdot
\frac{B_3A}{B_3C}\cdot
\frac{UC}{UD}
=
-
\frac{a}{c}\cdot
\frac{c}{b}\cdot
\frac{UC}{UD}\ .
\qquad\text{ So }
\frac{UC}{UD} = -\frac ba\ .
$$
We need the position of $C_3$ on $BD$. From $\frac{C_3A}{C_3B}=\frac{B_1A}{B_1C}=\frac{BA}{BC}=\frac ca$ we have $C_3A=c^2/(a+c)$, $C_3B=ac/(a+c)$. This gives $C_3D=C_3A+AD=C_3A+(a-c)=a^2/(a+c)$.
We are now in position to apply the reciprocal of Ceva in $\Delta DBC$ for the points $A_1$, $U$, $C_3$, so we compute:
$$
\frac{A_1B}{A_1C}\cdot
\frac{UC}{UD}\cdot
\frac{C_3D}{C_3B}
=
-
\frac cb\cdot
\frac ba\cdot
\frac {a^2/(a+c)}{ac/(a+c)}
=-1
\ .
$$
So $A_1D$ passes through $BU\cap CC3=X$.
This shows the concurrence of the six lines $AA_3$, $BB_3$, $CC_3$; $A_1D$, $B_1E$, $C_1F$.
$\square$
