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.

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

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:

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

• oh sorry my bad AF is angle bisector – Ualibek Nurgulan May 27 at 7:53
• Hello how did you draw the picture. What kinf of application di you use – Ualibek Nurgulan May 27 at 8:03
• @Ualibek Nurgulan: Yes, with $AF$ as a bisector it looks OK. I added a corrected image to he question. – g.kov May 27 at 8:14
• @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. – g.kov May 27 at 8:17

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 $$\angle \, ADI = 90^{\circ}$$.

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.