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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.

enter image description here

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    $\begingroup$ Where exactly is the point $F$ on the angle bisector of $\angle B$ $\endgroup$ – Anas A. Ibrahim May 27 at 6:19
  • $\begingroup$ It is AF. I mean AF is angle bisector my bad sorry. F is at BC $\endgroup$ – Ualibek Nurgulan May 27 at 7:56
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Something is definitely wrong with the question.

Checking out:

  1. $\triangle ABC$ be a scalene triangle.

  2. $I$ is incenter.

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

  4. $BF$ is angle bisector.

  5. Circumcircles of $\triangle ABC$ and $\triangle AEF$ meet at $A$ and $D$.

The image:

enter image description here

clearly illustrate that $\angle ADI$ is not anywhere near $90^\circ$.

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  • $\begingroup$ oh sorry my bad AF is angle bisector $\endgroup$ – Ualibek Nurgulan May 27 at 7:53
  • $\begingroup$ Hello how did you draw the picture. What kinf of application di you use $\endgroup$ – Ualibek Nurgulan May 27 at 8:03
  • $\begingroup$ @Ualibek Nurgulan: Yes, with $AF$ as a bisector it looks OK. I added a corrected image to he question. $\endgroup$ – g.kov May 27 at 8:14
  • $\begingroup$ @Ualibek Nurgulan:the picture was prepared using Asymptote, "Asymptote is a powerful descriptive vector graphics language that provides a natural coordinate-based framework for technical drawing. Labels and equations are typeset with LaTeX, for high-quality PostScript output." Also, you can check out this forum. $\endgroup$ – g.kov May 27 at 8:17
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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 $\angle \, ADI = 90^{\circ}$.

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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 $\odot(AXZIY)=\odot(S)$.

math stackexchange problem 3693462

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.

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