Prove that $U$ lies on the perpendicular bisector of $AI$ using cross ratios

Question

Let $I$ and $O$ be the incenter and circumcenter of $\triangle ABC$. Let $D$ be on $BC$ such that $ID \perp BC$. Let $E = AO \cap BC$, and let $T$ be a point such that $AT \perp BC$ and $IT \parallel BC$. Let $U = EI \cap DT$. Prove that $AU=UI$.

This problem was from a lecture about cross ratios, so ideally, we should be able to solve this with cross ratios. However, I can't find any four collinear points other than $B, D, E, C$, which gives nothing useful.

Can someone help? Thanks.

Answer 1

Let me add another approach:

As above, say $F=AT\cap BC$ is the projection of $A$ on $BC$. Let also $K=DI\cap AO$ and note $m(\angle KIA)=m(\angle IAF)=m(\angle KAI)$ since, as above $AI$ also bisects $\angle TAO$. Hence $KA=KI$ and to show $UA=UI$ it suffices to show $KU$ passes through the midpoint of $AI.$

Then, if $L\in EI$ is such that $KL$ is the internal bisector of $\angle DKE$ - i.e. $KL$ is parallel to $AI$ - to show $KU$ passes through the midpoint of $AI$ it suffices to show $(KA, KI: KU, KL)$ is a harmonic bundle so moving everything on the line $UE$ we want to show the cross-ratio $(U,L:I,E)$ is harmonic.

Now consider the intersection $P=TE\cap ID$ and recall the following fact: if $XYZT$ is any quadrilateral with $W=XZ\cap YT$ the intersection of the diagonals, and $V=XY\cap ZT$ and $R,S$ are the points where $WV$ meets $XT$ and respectively $YZ$ the the cross-ratio $(W,V:R,S)$ is always harmonic.

In our case, consider the quadrilateral $TDEI$ with $P$ playing the role of $W$ and $U$ playing the role of $V$. Since $TI$ and $DE$ are parallel it suffices to show $PL$ and $DE$ are themselves parallel and the conclusion follows from the above result.

Now, we see that $PL \parallel DE$ iff $\frac{IL}{LE} = \frac{IP}{PD}$. However, $$\frac{IL}{LE} = \frac{KI}{KE} = \frac{KA}{KE},$$where the first equality is from the angle bisector theorem and the second equality is from $KI=KA$. We also have $$\frac{IP}{PD} = \frac{TI}{DE} = \frac{FD}{DE},$$where the first equality is from $\triangle TIP \sim \triangle EDP$ and the second equality is from $KD \parallel AF$. Therefore, it suffices to prove that $\frac{KA}{KE} = \frac{FD}{DE}$, which is evident because $KD \parallel FA$.

Answer 2

The orthocenter and the circumcenter are isogonal conjugates, so $AI$ bisects $\widehat{TAE}$ and the circle centered at $I$ through $T$ is tangent to $AE$ at $S$. The radius of this circle is $BD-BF=(s-b)-c\cos B=(s-a)\frac{b-c}{a}$. Since $TF=ID=r$, $TD$ meets $AI$ at the midpoint $M$ of the $BC$-arc of the circumcircle, which is also a point on the perpendicular bisector of $BC$.

Let $F$ be the projection of $A$ on $BC$ and let us consider a dilation centered at $I$ with factor $2$. Since the midpoint of $TD$ is also the midpoint of $IF$, the $TD$ line is mapped into the $I_A F$ line, with $I_A$ being the $A$-excenter of $ABC$.

Let $G=I_A F\cap I_B I_C$: if we prove that $G,I,E$ are collinear we are done. We may notice that both $(AG,AB,AI,AC)$ and $(AG,AF,AI,AE)$ are harmonic bundles. Then $(AG\cap BC,F,AI\cap BC,E)$, $(AG\cap BC,I_B,A,I_C)$, $(AG\cap BC,G,A,I_A E\cap AG)$ are all harmonic bundles. By considering the perspectivity between the $I_B I_C$ and $BC$ lines centered ad $I$ we have that $G,I,E$ have to be collinear.