Prove that A₁D, B₁E, C₁F intersect at the same point

Question

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.

Answer 1

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.

Answer 2

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.

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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.)

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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.

Answer 3

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 $X$.

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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 $AA_3$, $BB_3$, $CC_3$; $A_1D$, $B_1E$, $C_1F$.

$\square$