Geometry problem proving that all lines $DE$ passes through one point

Question

Let $I$ is the incenter of $\triangle ABC$. Let $K$ be the circumcircle of $ABC$. Let $D$ be a variable point on arc $AB$ on $K$ not containing $C$. Let $E$ be a point on line segment $BC$ such that $\angle ADI = \angle IEC$. Prove that as $D$ varies on arc $AB$, the line $DE$ passes through one point.

First after testing a couple cases i found the point is the midpoint of arc $BC$. I've tried direct proof and reverse reconstructing the problem, but nothing worked. Can anyone help? (edit: I reflected $E'$ in the bisector of $C$ and created a cyclic quadrilateral $ADIE'$. Maybe it is a useful property...) (edit 2: I suspect inversion might be involved...)

Answer 1

Let $F$ be on parallel to $BC$ through $I$ so that $AF||IE$, then $\angle AFI = \angle IEC$.

Lemma: Line $EF$ goes through fixed point which is a midpoint of arc $BC$ not containing $A$.

Proof: Let line $EF$ meet $AI$ at $P$. Use Thales theorem twice. We have $${a\over b+p} = {EF\over EP} = {b\over p}\implies p ={b^2\over a-b}$$ so $P$ is fixed point on $AI$.

Now we try to identify $P$. Put $E$ in $B$. We see that $$\angle AFI = \angle IEC = \angle IBC= \angle IBA$$ so $AFE(B)I$ is concyclic. Now also see $$\angle CAP = \angle IAB = \angle BFI = \angle PBC$$ so $P$ lies also on circumcircle $ABC$ and the lemma is proven.

Answer 2

Comment:

As can be seen AD is always tangent to a circle center on I. Due to assumption the ratio of radius of inscribed circle (IG=r) and radii of these circles is constant. Consider position D' . $\angle D'AI=90^o$, E' is intersection of diameter D'P and BC. Let the foot of altitude from I on BC be G. Segments AI and IG are opposite to equal angles AD'I and IEC. The measures of AI and IG are constant . In this position DE(D'E') is coincident on diameter D'P and crosses midpoint P of arc BC. In position where D and E are coincident on B clearly $\angle ADI(=ABI) = \angle IEC(=IBC)$ because IB is bisector of ABC. In position where D is between A and B , ID and IE are parallel with BC and AD respectively, So $\angle ADE=\angle IEF$. The segments opposite to angles ADI and IEF are AI and IF. The measure of these two segments are constant.Now for three positions along arc AB the statement is true. This deduces that $\angle IEC=\angle ADI$ for all points along arc AB, because it is true for three positions. This is competent with our assumption.

Answer 3

Here is a detailed proof, also using "the point $F$" as in the post of Aqua, it is the point that best "geometrically interpolates" between $D$ and $E$. The definition of $F$ below will be the one, that lets us give a direct (instead of indirect) proof.

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We start by constructing the parallel $(f)$ to $BC$ through $I$. Then let us draw the circle $(\Gamma)$ through $A,D,I$, it intersects $(f)$ first time $I$, and for a second time (counting multiplicities) in a point $F$. This is the definition of $F$. By construction, we have the same measure $x$ of angles... $$ x =\widehat{IEC} =\widehat{ADI} =\widehat{AFI} \ . $$ The points $A'$, and $S$ are the intersections of the angle bisector $AI$ (in $A$ of $\Delta ABC$) with the side $BC$, and respectively with the circumcenter $(K)=\odot(ABC)$.

Claim:

• $(1)$ The points $S,D,F$ are colinear.
• $(2)$ The points $S,E,F$ are colinear.

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It is clear, that the OP follows from the above, showing that $DE$ passes through the fixed point $S$, the mid point of the arc $\overset\frown{BC}$ opposite to the vertex $A$.

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Proof of the claim:

$(1)$ Using the two circles, the given $(K)$ and the constructed $(\Gamma)$ we have the equality of measures of angles: $$ \begin{aligned} \widehat{ADS} &\overset{(K)}= \widehat{ABS} = \widehat{ABC}+\widehat{CBS} = \hat B+\widehat{CAS} = \hat B+\frac 12\hat A\ , \\ \widehat{ADF} &\overset{(\Gamma)}= \widehat{FIS} = \widehat{BA'S} = \frac12(\overset\frown{BS}+\overset\frown{AC}) = \frac12\overset\frown{BS}+\frac12\overset\frown{AC} = \widehat{BAS}+\widehat{ABC} = \frac 12\hat A+\hat B\ , \\ &\qquad\text{ giving } \\ \widehat{ADS} &= \widehat{ADF} \ , \end{aligned} $$ so the points $D,F,S$ are colinear. This concludes $(1)$.

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$(2)$ We will show that $F,E,S$ are colinear by the reciprocal of the theorem of Thales, showing that proportions of corresponding sides in the a posteriori valid similarity $\Delta SA'E\sim\Delta SI$ are equal. Here is the computational path: $$ \begin{aligned} \frac {FI}{EA'} &= \frac {FI}{EA'} &&\text{ using }\Delta AIF\sim\Delta IA'E \\ &= \frac {AB}{BA'} &&\text{ using the angle bisector $BI$ in}\Delta ABA' \\ &= \frac {SC}{SA'} &&\text{ using }\Delta SA'C\sim\Delta BA'A \\ &= \frac {SI}{SA'} &&\text{ using $\Delta SIC$ isosceles, $SC=SI$} \ . \end{aligned} $$ The needed proportions are thus equal, so $F,E,S$ are colinear. This concludes $(2)$, and the claim.