###
**1. Introduction**

The following property of mixtilinear incircle is quite-known**Theorem 1.$[1]$**Let

**$(I)$ be incircle of triangle $ABC$. $P$ is an arbitrary point on the circumcircle of triangle $ABC$. Two tangent lines through $P$ of the incircle $(I)$ intersect $BC$ at $X$, $Y$.**

Then the circle $PXY$ passes through the common point of $A-$mixtilinear incircle of triangle $ABC$ and the circumcircle of triangle $ABC$.

It was generalized by

**Vladimir Zajic**(nick name

*yetti*on AoPS) as follow

**Theorem 2. $[2]$**Let two tangent lines of the incircle of triangle $ABC$ meet at $X$ and intersect $BC$ at $Y$, $Z$. $AX$ intersects the circumcircle of triangle $ABC$ at $K$.

Then the circle $KYZ$ passes through the common point of $A-$mixtilinear incircle of triangle $ABC$ and the circumcircle of triangle $ABC$.

### 2. Generalization and Proof

Recently,

**Nguyen Duc Bao**(nick name*baopbc*on Vietnam Mathematics Forum) found a generalization of mixtilinear circle as follow**Theorem 3.$[3]$**$P$, $Q$ are isogonal conjugate points with respect to triangle $ABC$.

Tangent line at $P$ of the circle $PBC$ intersects $AB$, $AC$ at $P_{ab}$, $P_{ac}$, respectively.

Tangent line at $Q$ of the circle $QBC$ intersects $AB$, $AC$ at $Q_{ab}$, $Q_{ac}$, respectively.

Then $P_{ab}$, $P_{ac}$, $Q_{ab}$, $Q_{ac}$ lie on a circle, which is tangent to the circumcircle of triangle $ABC$.

Proof of theorem 3 can easily obtained by angle chasing and the fact:

*the tangency point is Miquel point of the quadrilateral $(PB,PC,QB,QC)$.*

Independently,

**Telv Cohl**has also found this generalization with various of interesting properties in $[4]$.

Let $T$ be the tangency point in theorem 3. Then we generalized theorem 2 as follow

**Problem 4.**Let $\mathcal{S}$ be the inconic of triangle $ABC$ of which foci are $P$, $Q$.

Two tangent lines of $\mathcal{S}$ meet at $X$ and intersect $BC$ at $Y$, $Z$. $AX$ intersects the circumcircle of triangle $ABC$ at $K$.

Prove that the circle $KYZ$ passes through $T$.

**Proof.**Thanks to the proof of

**Luis Gonzales**for theorem 2 and

**Telv Cohl's results$[4]$,**I have completed a synthetic solution.

Follow

**Telv Cohl's proof for property 2$[4]$,**if we let $A'$ be a point on the circumcircle of triangle $ABC$ such that $AA'$ is parallel to $BC$ then $T$, $D$, $A'$ are collinear.

Let $\mathcal{S}$ touches $BC$ at $D$ and $AK$ intersects $BC$ at $D'$.

According to Desargues's involution theorem, there exists a point $R$ on $BC$ such that

$RB\cdot RC=RD\cdot RD'=RY\cdot RZ$

Simply by angle chasing, $(TD,TK)=(TA',TK)=(AA',AK)=(BC,AK)=(D'D,D'K)$. Thus $T$, $K$, $D$, $D'$ are concyclic. Combine with $T$, $K$, $B$, $C$ are concylic, $TK$ passes through $R$.
$\Rightarrow RY\cdot RZ=RK\cdot RT$

Hence we conclude that $K$, $Y$, $Z$, $T$ are concyclic.By the inversion centered at $T$, we obtain that

**Corollary 5.**$P'$, $Q'$ are reflections of $P$, $Q$ in the circumcircle of triangle $ABC$.

Then $T$, $A$, $P$, $Q'$ are concyclic and $T$, $A$, $Q$, $P'$ are concyclic.

Thanks anh nhieu, chuc anh suc khoe

Trả lờiXóadang ky tai khoan bong88, tai khoan dung thu bong88, tai khoan dung thu bong88