Triangle, incenter and perpendicular line

Question

For the triangle $ABC$, let $I$ the incenter and $I_A$ the A-excenter. If $L$ the midpoint of arc $BC$, we can show that $L$ is the center of a circle through $I, I_A, B, C.$ Also, if the incircle touches $AB, AC$ at $P, Q$ and $BI, CI$ intersect with $PQ$ at $K, L$ we may show that circumcircle of $ILK$ is tangent to incircle of $ABC$ if and only if $AB+AC=3BC.$ My problem is that if the line through $I$ is perpendicular to $BI$ and meets $AC$ at $X,$ while the line through $I$ perpendicular to $CI$ meets $AB$ at $Y$ and $AB+AC=3BC$, I stuck to show that $X, I_A, Y$ are collinear

Answer 1

It is hard to extract a question from the OP. So i suppose the question is solved by the following result:

Let $\Delta ABC$ be a triangle with sides $a,b,c$. Let $p=(a+b+c)/2$ be its half-perimeter. We assume $$ 3a=b+c\ . $$ Let $I$ and $I_A$ be the incenter and the $A$-excenter of $\Delta ABC$. Let $P\in AB$, $Q\in AC$, $R\in BC$ be the touching points of the inscribed circle $(I)$ (centered in $I$).

• Then $AP=AQ=p-a=BC$.

Let $W$ be the mid point of $AI$.

Let $RI$ further intersect the circle $(I)$ in a point $U$ (with $U\ne R$). Parallel and equal to the segment $IR$, we construct in the half plane delimited by $BC$, which does not contain $I,A$, the segments $BB^*$, and $CC^*$. These segments are in particular perpendicular on $BC$, and the quadrilaterals $UIB^*B$, $IRB^*B$; $UIC^*C$, $IRC^*C$ are parallelograms. Let now $V$ be the second point on the circumcircle $(ABC)$, where the angle bisector from $A$ intersects this circle. ($V\ne A$.) It is characterized by
$$\widehat{VBC}=\widehat{VCB}=\frac 12\hat A\ ,$$ and that is located on the angle bisector $AII_A$, and its intersection with the side bisector of $BC$ (through $R'$, the mid point of $BC$).

• Show that the triangles $\Delta AIP$, $\Delta AIQ$, $\Delta BC^*C$, $\Delta CB^*B$ are congruent.

• Show that $VCAB$ is a cyclic quadrilateral. Show that $BCC^*B^*$ is a rectangle, and its center $BC^*\cap B^*C$ is $V$. In particular, the hexagon $ICC^*I_AB^*B$ is cyclic, the center of the circle $(ICC^*I_AB^*B)$ coincides with $V$, and we have moreover its radius: $$ VI=VB=VB^*=VI_A=VC^*=VC=\frac 12BC^*=\frac 12B^*C=\frac 12AI=AW=WI\ . $$

We define two points $K,L$ after defining $U$ as follows. Consider the triangle $UBC$. Its heights are $UR$, and $BK,CL$, thus defining $K\in UC$, $L\in UB$.

• Show that $I$ is the orhtocenter of $UBC$, $$I=UR\cap BK\cap CL\ . $$

• Show that $IKUL$ is a cyclic quadrilateral. The segment $UI$ is a diameter of the circle $(IKUL)$.

• Show that $P,L,K,Q$ are colinear.

The parallel through $I$ to the line $ULB$ intersects $AB$ in a point denoted by $Y$.

The parallel through $I$ to the line $UKC$ intersects $AC$ in a point denoted by $X$.

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• Show that $X,I_A,Y$ are colinear.

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Proof:

• The relation $AP=AQ=p-a$ holds in a general triangle. In our case, we have further $p-a=(a+b+c)/2-a=(a+3a)/2-a=2a-a=a=BC$.

• The triangles $\Delta AIP$, $\Delta AIQ$, $\Delta BC^*C$, $\Delta CB^*B$ are congruent, because they have a right angle in $P$, $Q$, $C$, and respectively $B$, and they have equal catetes, $a=AP=AQ=BC$, and $IP=IQ=IR=BB^*=CC^*$. In particular, the hypotenuses are correspondingly equal, and the other angles, too: $$ \begin{aligned} &AI = BC^*=B^*C\ ,\\ &\widehat{C^*BC}=\widehat{B^*CB}=\frac 12\hat A \ . \end{aligned} $$ By construction, $BB^*$, $CC^*$ are parallel, equal, and build right angles with $BC$. So $BCC^*B^*$ rectangle. The intersection of the diagonals satisfies the above property that determines $V$, $\widehat{C^*BC}=\widehat{B^*CB}=\frac 12\hat A $, so this intersection is on the circle $(ABC)$, and it is the mid point of the arc $\overset \frown{BC}$, so it is $V$. The quadrilateral $IBI_AC$ is also cyclic, two opposite right angles, and $II_A$ is a diameter of the circle $(IBI_AC)$. Its mid point is $V$, the second intersection point of $AII_A$ with the circle $(ABC)$, so the two circles $(IBI_AC)$ and $(BCC^*B^*)$ coincide, and we have the long equality of segments: $$ AW=WI=IV=VB=VC=VB^*=VC^*=VI_A\ . $$

• Let us now show the properties related to $U$. By construction, the segment $UI$ is parallel and equal to $BB^*$ and $CC^*$. So $UBB^*I$ is a parallelogram. So $$ \widehat{BUR}= \widehat{BUI}= \widehat{BB^*I}= \widehat{BCI}\ . $$ The line $CI$ is thus perpendicular on $BU$. Similarly $BI\perp CU$. So $I$ is the orthocenter in $UBC$, and $B,I,K$ are colinear, and $C,I,L$ are colinear.

• Let us show that $P,Q,K,L$ are colinear. For this, we use an inversion $\#$ with center $I$ and power $-IP^2=-IQ^2=-IR^2$, this argument is simplest from my point of view, a picture for this argument would be: The point $P$ is transformed in its opposite $P^{\#}$ w.r.t. the center of inversion $I$, so $PP^{\#}$ is a diameter in $(I)$. The point $Q$ is transformed in its opposite $Q^{\#}$, so $QQ^{\#}$ is a diameter in $(I)$. The point $K$ is transformed in $K^{\#}=B$. (Because the triangles $\Delta IKU$, $\Delta IRB$ are similar, so $IK\cdot IB=IU\cdot IR=-IR^2$.) Similarly, $L$ is transformed in $L^{\#}=C$. To show that $P,Q,K,L$ are colinear, we show equivalently that the transformed points $P^{\#}$, $Q^{\#}$, $B=K^{\#}$, $C=L^{\#}$ are on a circle through $I$. (This is of course the circle $(IBC)$, and we already have a lot of points on it, i wanted two more, this is the reason for this approach.) It is enough to show $P^{\#}$ on the circle $(IBC)$. We show this by comparison of the angles in $P^{\#}$ and $I$ in the quadrilateral $IP^{\#}CB$. A first progress in this direction is given by the remark that $U,K,P^{\#},C$ are colinear. (Because their transformed points $R,B,P,L$ are on a circle through $I$, yes, it is a circle with diameter $BI$.) So $P^{\#}$ is the symmetric of $U$ w.r.t. $K$. We then compute: $$ \begin{aligned} \widehat{BP^{\#}C} &= \pi-\widehat{BP^{\#}U}\\ &= \pi-\widehat{BUP^{\#}}\\ &= \pi-\widehat{LUK}\\ &= \widehat{LIK}\\ &= \widehat{BIC}\ . \end{aligned} $$ The comparison shows we have to equal angles in $P^{\#}$ and $I$ in the quadrilateral $IP^{\#}CB$, so it is cyclic, this is exactly what we wanted. Same game for $Q^{\#}$.

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• We finally attack the question from the OP.

Let us show the colinearity of $X,Y,I_A$. (I could not find a simple Desargue or Pappus argument, mainly because the points $X,Y$ are on lines defined by too few "distinguished points". So the argument is inversion again. So i inserted the above "short prelude", same thematical idea.) We denote by $X^{\#}$, $Y^{\#}$, $I_A^{\#}$ the points obtained from $X,Y,I_A$ by the same inversion $\#$.

Firstly, $I_A^{\#}$ is the point on $II_A$ at the intersection with $PQ$ because of $$ II_A^{\#}\cdot IA = IP^2\ . $$ Secondly, the point $Y^{\#}$ is on $IB^*$, and also on the circle which is the image by inversion of the line $APB$. The point $A^{\#}$ is the intersection of $IA$ with $P^{\#}Q^{\#}$. It is also the symmetrical point of $I_A^{\#}$ w.r.t. $I$. The points $P,Q$ have already discussed inverses, $P^{\#}$, $Q^{\#}$, so that $PP^{\#}$, and $QQ^{\#}$ are diameters. Shortly, $Y^{\#}$ is the point on the circle with diameter $IP^{\#}$ which is the second intersection point with the line perpendicular on $IL$ in $I$.

The point $X^{\#}$ is located similarly, it is the point on the circle with diameter $IQ^{\#}$ which is the second intersection point with the line perpendicular on $IK$ in $I$.

We want to show that the points $I, X^{\#}, Y^{\#},I_A^{\#}$ are on a circle.

We construct the point $Z$ as the intersection $Q^{\#}X^{\#}\cap P{\#}Y{\#}$.

Observe that $Q^{\#}X^{\#}\|IK$, both lines being perpendicular on $X^{\#}IC^*$. So $Q^{\#}X^{\#}Z\perp UP^{\#}$, because $IK\perp UP^{\#}$. Similarly $P^{\#}Y^{\#}Z\perp UQ^{\#}$.

So $Z$ is the orthocenter in the triangle $\Delta UQ^{\#}P^{\#}$.

So $UZ\perp Q^{\#}P^{\#}$.

So $UZ\perp PLKQ$. We want to show that $Z$ is on this last line. For this observe that $\Delta UPQ\equiv \Delta ZQ^{\#}P^{\#}$, because the sides are parallel, and one corresponding side has the same length,
$PQ\equiv Q^{\#}P^{\#}$. So $UPQ^{\#}Z$ and $UQP^{\#}Z$ are parallelograms. So $UZ=PQ^{\#}=QP^{\#}$. But $PLKQ$ is mid line in $\Delta UQ^{\#}Q^{\#}$, so the distance from $U$ to $PLKQ$ is the same as the distance $PQ^{\#}=QP^{\#}$ between the parallels $PQ\| Q^{\#}P^{\#}$. So $Z$ is on the mid line.

From $II_A^{\#}A\perp PLKQ$ we immediately conclude:

The points $X^{\#}$, $Y^{\#}$, $I_A^{\#}$, lie on the circle with diameter $IZ$.

$\square$