Let $ABC$ be a triangle such that $\angle ABC = 90 ^{\circ}$. Let $I$ be an incenter of $ABC$ and let $F$, $D$ and $E$ be points where incircle touches sides $AB$, $BC$ and $AC$, respectively. If lines $CI$ and $EF$ intersect at point $M$ and if $DM$ and $AB$ intersect in $N$, prove that $AI=ND$
Problem
Source: Bosnia and Herzegovina Junior Balkan Mathematical Olympiad TST 2017
Tags: geometry, incircle, right angle, incenter
amar_04
23.08.2019 01:30
Shocked to see that no one has posted a solution yet.. We will use a very well known problem (or you can call it as a lemma) Lemma: $ABC$ is a triangle and if the incircle touches $BC,CA,AB$ at $D,E,F$ and respectively and $I$ is the incenter of $\triangle ABC$ and if $CI\cap FE=M$, then $BM\perp CM$. Proof of the lemma: Join $I$ to $F$. Let $\angle BAC=\theta\implies\angle MFB=90^\circ+\frac{\theta}{2}$ also $\angle DIM=90^\circ-\frac{\theta}{2}\implies BIMF$ is a cyclic quadrilateral. So, $\angle BMC=\angle BFI=90^\circ$. Main Proof: Notice that $BFID, BDIM$ are cyclic quadrilaterals. Hence $FMID$ is also a cyclic quadrilateral $\implies\angle FMD=\angle FID=90^\circ$. Let $AI\cap EF=K$. So, $\angle FKA=90^\circ\implies ND\| AI$. Now it's angle chasing time... $\angle NDI=\angle MFI=\angle FND=180^\circ-\angle DNA\implies AN\| DI$. So, $NDIA$ is a parallelogram. Hence, $DN=IA$. $\blacksquare$.
Milana
29.07.2020 16:09
amar_04 wrote: Shocked to see that no one has posted a solution yet.. We will use a very well known problem (or you can call it as a lemma) Lemma: $ABC$ is a triangle and if the incircle touches $BC,CA,AB$ at $D,E,F$ and respectively and $I$ is the incenter of $\triangle ABC$ and if $CI\cap FE=M$, then $BM\perp CM$. Proof of the lemma: Join $I$ to $F$. Let $\angle BAC=\theta\implies\angle MFB=90^\circ+\frac{\theta}{2}$ also $\angle DIM=90^\circ-\frac{\theta}{2}\implies BIMF$ is a cyclic quadrilateral. So, $\angle BMC=\angle BFI=90^\circ$. Main Proof: Notice that $BFID, BDIM$ are cyclic quadrilaterals. Hence $FMID$ is also a cyclic quadrilateral $\implies\angle FMD=\angle FID=90^\circ$. Let $AI\cap EF=K$. So, $\angle FKA=90^\circ\implies ND\| AI$. Now it's angle chasing time... $\angle NDI=\angle MFI=\angle FND=180^\circ-\angle DNA\implies AN\| DI$. So, $NDIA$ is a parallelogram. Hence, $DN=IA$. $\blacksquare$. How did you get $\angle DIM=90^\circ-\frac{\theta}{2}$?
rafigamath
25.05.2023 21:09
We will prove that $ANDI$ is a parallelogram. First notice that $AN\perp BC, ID\perp BC$ $\implies AN\parallel ID$. Since $AF=AE$ and $AI$ angle bisector,$AI\perp FE$ . Let $\angle{ICB}=\alpha$. Then $\angle{AEF}=45+\alpha$ and $\angle{AEM}$ is the external angle,so $\angle{EMC}= 45$ which implies $BFMID$ cyclic. So $\angle{FBD}=\angle{DME}=90$,$ND\perp FE$ and we're done