Points $D$ and $E$ lie on the lateral sides $AB$ and $BC$ respectively of an isosceles triangle $ABC$ in such a way that $\angle BED = 3\angle BDE$. Let $D'$ be the reflection of $D$ about $AC$. Prove that the line $D'E$ passes through the incenter of $ABC$.
Problem
Source: Sharygin 2023 - P4 (Grade-8)
Tags: geometry, incenter, Sharygin Geometry Olympiad, Sharygin 2023
Tintarn
04.03.2023 11:22
If $D$ moves linearly on $AB$, then also $D'$ and $E$ move linearly. So it suffices to consider the two extremal cases $D=A$ and $D=B$.
The latter is trivial since then $E=B$ and $D'E$ is the angular bisector of $\angle CBA$, hence passing through $I$.
If instead $D=A$, then also $D'=A$ and the condition on the angles shows that $D'E=AI$ is the angular bisector of $\angle BAC$; hence again passing through $I$.
gghx
07.03.2023 12:37
Let the angle bisector of $\angle BAC$ intersect $D'E$ at $I$ and $BC$ at $K$. Let $\angle BDE=x$, by angle chase $\angle BED=3x$, $\angle DBE=180-4x$, $\angle BAC=2x$, $\angle BAK=x\implies AK\parallel DE$, $\angle DAD'=4x\implies AD'\parallel BC$. Then $$\frac{AI}{IK}=\frac{AD'}{EK}=\frac{AD}{EK}=\frac{BA}{BK}=\frac{AC}{CK}\implies CI\text{ bisects }\angle BCA.$$
sunken rock
08.03.2023 10:48
Let $AF$ be bisector of $\widehat{BAC}=2\alpha, F\in BC\implies\widehat{AFB}=3\alpha\implies DE\parallel AF$. Take $B'$ reflection of $B$ about $AC, BB'\parallel DD'$ and: $\frac{BE}{EF}=\frac{BD}{AD}=\frac{B'D'}{AD'}$, which proves that $AF,BB',ED'$ are concurrent, $AF$ and $BB'$ being angle bisectors, we are done. Best regards, sunken rock
Attachments:
PROF65
08.03.2023 15:22
General Problem Let $ABC$ be a triangle with incenter $I$ $D,E$ two points of $AB ,BC$ s.t. $DE\parallel AI $ $D'$ is the intersection of the parallel to $BC$ through $A$ and the parallel to $ BI $ through $ D$. Prove that $E,I,D'$ are collinear. proof : Let $d_1$ be the infinity point of $AI$; $d_2$ be the infinity point of $BI$; Let $d_3$ be the infinity point of $BC$ applying Pappus's to $ \binom{B\ D \ A}{ d_1\ d_3\ d_2}$ we get $E,I,D'$ are collinear . Back to the problem $\angle BDE= \frac{\angle A+\angle C}{4}=\angle BAI$ thus $DE\parallel AI$ besides $DD' \parallel BI$ so applying the proved 'general problem' we get the desired result . Best regards RH HAS
djmathman
13.03.2023 00:30
Tintarn wrote: If $D$ moves linearly on $AB$, then also $D'$ and $E$ move linearly. So it suffices to consider the two extremal cases $D=A$ and $D=B$. I might be missing something, but I think you explicitly need the fact that the lines $D'$ and $E$ move on are parallel to each other (which fortunately is true in this case). Otherwise, all such lines $D'E$ are not guaranteed to pass through a fixed point.