Let $ABC$ be a triangle with incenter $I$. The points $D$ and $E$ lie on the segments $CA$ and $BC$ respectively, such that $CD = CE$. Let $F$ be a point on the segment $CD$. Prove that the quadrilateral $ABEF$ is circumscribable if and only if the quadrilateral $DIEF$ is cyclic. Proposed by Dorlir Ahmeti, Albania
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Tags: geometry
13.09.2020 00:40
We will repeatedly use the fact that a quadrilateral $ABCD$ is cicumscribable iff the angle bisectors of $\angle ABC,\angle BCD,\angle CDA, \angle DAB$ are concurrent at a point. Note that $CD=CE$ implies that $D,E$ are symmetrical points wrt $CI$. Suppose $ABEF$ a incircle. So $FI,EI$ bisects $\angle AFE , \angle FEB$ respectively. So we have: $$ \angle FEI = \angle IEB= \angle IDA = \pi - \angle CDI = \pi - \angle FDI \implies \{FDIE\} \quad \text {is cyclic}$$ Now suppose $\{FDIE\}$ is cyclic. Suppose $F'\in AB$ is the reflection of $F$ in $CI$. Note that since $ID=IE$ , hence by Fact 5, $FI$ bisects $\angle DFE$ . Also we have : $$\angle IEB=\pi- \angle IEF'=\angle IDF'= \angle IEF$$ This shows that $EI$ bisects $\angle FEB$ and we are done.
13.09.2020 00:57
Aryan-23 wrote: I think I really like this problem . Although not very difficult, I really found the problem statement very attractive. It would have been a perfect problem for JBMO ! Thank you for your words. I did have hopes on this problem to be on the test since it has unique statment. It is not difficult but it is nice one.
23.06.2021 14:15
Do we need to prove that all the ancle bisectors of $ABEF$ pass through $I$, or it is enough to prove that 3 of them do???
08.07.2023 15:23
sttsmet wrote: Do we need to prove that all the ancle bisectors of $ABEF$ pass through $I$, or it is enough to prove that 3 of them do??? Proving 3 angle bisectors are concurrent in a quadrilateral is sufficient. You can prove this by taking the feet of perpendicular from the concurrent point, and show this implies all 4 angle bisectors are concurrent