Prove that $∡ADI=90°$ Let $ABC$ be a scalene triangle. $I$ is incenter. Common point of inscribed circle and $BC$ is $E$. $AF$ is angle bisector. If circumcircles of $ABC$ and $AEF$ meet at $A$ and $D$, then prove $∡ADI=90°$. MY TRY: I chased angle. And got little value of result. I have found out $∡CDE=∡EDB=∡BAF=∡FAC$. And other than this I can only tell $ABCD$ and $ADEF$ are circumcircle.

 A: Something is definitely wrong with the question.
Checking out:


*

*$\triangle ABC$ be a scalene triangle.

*$I$ is incenter.

*Common point of inscribed circle and $BC$ is $E$. 

*$BF$ is angle bisector.

*Circumcircles of $\triangle ABC$ and $\triangle AEF$ meet at $A$ and $D$.
The image:

clearly illustrate that 
$\angle ADI$ is not anywhere near $90^\circ$.
A: Denote by $k_0$ the circumscribed circle of triangle $ABC$ and by $k_1$ the circumcircle of triangle $AEF$, where by assumption $k_0 \cap k_1 = \{A, D\}$. Extend the angle bisector $AF$ until it intersects the circumcircle $k_0$ of $ABC$ into the second point $L$ on $k_0$, the first being $A$. Then $L$ is the midpoint of the arc of $k_0$ between the points $B$ and $C$ that does not contain point $A$, because $AL$ is the angle bisector of angle $\angle \, BAC$. Therefore $LB = LC$. After some very simple angle chasing, one can show that $\angle\, LBI = \angle\, LIB$, which means that the triangle $BLI$ is isosceles with $LB = LI$. Thus we have that $$LB = LC = LI$$
Construct the circle $\omega$ with center $L$ and radius $LB$. Then the three points $B, \, C,\, I$ lie on $\omega$. 
If you perform inversion with respect to $\omega$, the circle $k_0$ is mapped to the line $BC$ and in particular the point $A$ is mapped to the point $F$. However, both points $A$ and $F$ lie on circle $k_1$, which means that circle $k_1$ is mapped to itself under the inversion in $\omega$ (and is in fact orthogonal to $\omega$). Since $k_0 \cap k_1 = \{A, D\}$ their image under the inversion with respect to $\omega $ is $BC \cap k_1 = \{F, E\}$ which means that the point $E$ is mapped to the point $D$ under the inversion and the points $D, E$ and $L$ are collinear. 
Now, consider circle $k_2$ circumscribed around triangle $EFI$. Since $IE \, \perp \, BC$  we see that $\angle\, IEF = 90^{\circ}$, which means that the center $O_2$ of $k_2$ is the midpoint of segment $IF$ so $O_2$ lies on the angle bisector $AL$ and thus the points $L, \, O_2,\,  I$ are collinear. Hence circle $k_2$ is tangent to circle $\omega$ at point $I$. Under inversion in $\omega$, the circle $k_2$ is mapped to the circle $k_3$ passing through the image points $I, \, A, \, D$ of points $I, \, F, \, E$ respectively, and $k_3$ is also tangent to $\omega$ at point $I$. Hence the center of $k_3$ must be collinear with the the centers $L$ and $O_2$ of $\omega$ and $k_2$, which lie on the angle bisector $AL$, so the center of $k_3$ also lies on $AL$ and therefore the center of $k_3$ lies on the segment $AI$. The latter fact however means that $AI$ is a diameter of $k_3$. Since, as already established, $D$ lies on $k_3$, angle $\angle \, ADI = 90^{\circ}$.      
A: The problem in the OP can be stated equivalently in the form: The circles $\odot(ABC)$, $\odot(AEF)$, and the circle with diameter $AI$ have a common chord. (Which is $AD$ in the OP.) The first and the third in the list are "simpler" (for my taste, they depend on "simpler points"), so let $D'$ be their intersection, let us try to show that the third circle, $\odot(AEF)$ also passes through $D'$. (So $D=D'$ in the final.)
As it often happens in problems involving essentially the centers $O,I$ (of the circumcircle and the incircle) of a triangle, the following constellation of points is useful:
wiki page on Euler's formula
The following solution is based on the point $L$ from the above link (and the solution by inversion of Futurologist), and on the projection $Z$ of $I$ on the $A$-height. The idea of the following solution is to show that $L,E,Z,D'$ are collinear. 

In the following picture, let $AH$ be the height in $A$, $H\in BC$, let $X,Y,Z$ be the projections of $I$ on $AB$, $BC$, $AH$.
(So $EYX$ is the incircle.)
Let $S$ be the mid point of $AI$. Let $\odot(S)$ be the circle centered in $S$ with diameter $AI$.
Let $D'\ne A$ be the second intersection of the circles $\odot(ABC)=\odot(O)$ and $\odot(AXZIY)=\odot(S)$.

We have:
$$
\begin{aligned}
\widehat{LD'A}
&=
\widehat{LBA}
=
\widehat{LBC} + \widehat{CBA} = \frac 12 \hat A+\hat B\ ,
\\
\widehat{ZIA}
&=
\widehat{BFA}
=
\frac 12\overset{\frown}{AB}
+
\frac 12\overset{\frown}{LC}
=\hat C+\frac 12 A\ ,
\\
\widehat{ZD'A}
&=
180^\circ -\widehat{ZIA}
=180^\circ -\left(\hat C+\frac 12 A\right)
=\frac 12 \hat A+\hat B
=\widehat{LD'A}\ .
\end{aligned}
$$
So $L,Z,D'$ are on the same line. Let us show now that $E$ is also on this line. For this, we compute two proportions, this seems to be the quick+dirty path:
$$
\begin{aligned}
\frac{IE}{AZ}
&=
\frac{ZH}{AZ}
=
\frac{FI}{IA}
=
\frac{BF}{BA}
=
\frac{ac/(b+c)}c
=
\frac a{b+c}
\ ,\\[2mm]
\frac{LI}{IA}
&=
\frac{LB}{LA}
=
\frac{\sin\widehat{BAL}}{\sin\widehat{ABL}}
=
\frac{\sin(\hat A/2)}{\sin(\hat A/2+\hat B)}
=
\frac{2\sin(\hat A/2)\cos(\hat A/2)}{2\sin(\hat A/2+\hat B)\cos(\hat A/2)}
\\
&=
\frac{\sin\hat A}{\sin\hat B+\sin\hat C}
=\frac a{b+c}\ .
\end{aligned}
$$
So $\Delta LIE\sim\Delta LAZ$ (since the above proportions are equal, and the angles in $I$ and $A$ are correspondent), so they have the same angle in $L$, so $L,E,Z$ colinear.

We can now conclude. The quadrilaterals $ZIAD'$ and $EFAD'$ have the same angles, so the second one is also cyclic, as the first one. This means that $AD'$ is also a chord in $\odot(AD'XZIY)$. (So $D=D'$.)
$\square$

Bonus: With the notations from the picture, $S$ is on $MN$. The circle $\odot(S)$, and the lines $CI$, $EX$ intersect in a point. The circle $\odot(S)$, and the lines $BI$, $EY$ intersect in a point.
