Simple easy problem ; infact one can even show ce = cf as well .

$\Delta IFC$ and $\Delta IEC$ are congruent....easily proved by angle-chase

Very easy problem... Trivial angle chase.....

Sorry to bump this, but I don't understand how to do this problem. I've tried angle bisectors, angle chase, Pythagorean theorem, etc. but I can't show that $IF=IE$. Maybe I am drawing the diagram wrong. Can somebody please provide a correct diagram for this problem?

ThematicMania wrote: Sorry to bump this, but I don't understand how to do this problem. I've tried angle bisectors, angle chase, Pythagorean theorem, etc. but I can't show that $IF=IE$. Maybe I am drawing the diagram wrong. Can somebody please provide a correct diagram for this problem? We assume that $\angle BAC=\alpha$.Thus, $\angle BCA=90^\circ -\alpha$. From $AI \perp IE$ $\Rightarrow$ $\angle IEA=90^\circ-\dfrac{\alpha}{2}$. So,we have $\angle FCI=\angle ICE=45^\circ-\dfrac{\alpha}{2}.$. $\angle CIF=\angle CAI+\angle ICA=45^\circ$.Similarly, $\angle CIE=\angle IEA-\angle ICE= =90^\circ-\dfrac{\alpha}{2}-45^\circ+\dfrac{\alpha}{2}=45^\circ$ So, $\triangle CIF$ similar to $\triangle CIE$ $\Rightarrow IE=IF$.

Another Solution: Extend $IE \cap BC=D$, then $ADBI$ is cyclic, easy to see, $\angle IFC=90^{\circ}+\frac{A}{2}$ and, $\angle DEC=45^{\circ}+45^{\circ}+\frac{A}{2}$ $=$ $\angle IFC$, then finish off with congruency ($\Delta IFC \cong \Delta IEC$)

The first RMO geo which I solved myself Its easy to see that $\angle IEC=\angle AIE+\angle IAE=90+\frac{A}{2}$ and also, $\angle AFC=\angle BAF+\angle ABF=90+\frac{A}{2}$ Also, $\angle IFC=\angle ICE=\frac{C}{2}$ So, $\triangle IFC\cong\triangle IEC$, and thus, $IF=EF$ by CPCT Q.E.D.(rare moment for me) Thanks @below

@above congrats on your first RMO geo problem .

Let's draw $IP,IQ$ such that $IP \perp AB$ and $IQ \perp BC$. So we have $IP=IQ$. In $\triangle API $ and $\triangle AIE$ we have $\angle PAI =\angle IAE =\frac{A}{2}$ and obviously a $90$ in common . so $\triangle PAI \sim \triangle IAE$. So ,$\frac{AI}{AE}=\frac{PI }{IE} \cdots (1)$. Similarly we have $\triangle IQF \sim \triangle IAE$. So,$\frac{IF}{AE} =\frac{IQ}{AI} \cdots (2)$. Comparing $(1)$and$(2)$ get, $\boxed{IE=IF}$.

ftheftics wrote: Let's draw $IP,IQ$ such that $IP \perp AB$ and $IQ \perp BC$. So we have $IP=IQ$. In $\triangle API $ and $\triangle AIE$ we have $\angle PAI =\angle IAE =\frac{A}{2}$ and obviously a $90$ in common . so $\triangle PAI \sim \triangle IAE$. So ,$\frac{AI}{AE}=\frac{PI }{IE} \cdots (1)$. Similarly we have $\triangle IQF \sim \triangle IAE$. So,$\frac{IF}{AE} =\frac{IQ}{AI} \cdots (2)$. Comparing $(1)$and$(2)$ get, $\boxed{IE=IF}$. nice solution

I have another solution which also uses some construction (but something other than the official solution). So here goes my solution to the problem : Let line IE intersect segment AB at G. As triangle As triangle AIE is congruent to triangle AIG, therefore IE = IG . Now since quadrilateral BGIF is cyclic, so ∠IFG = ∠IBG = ∠B/2 = ∠IBF = ∠IGF, therefore IG = IF. So IE = IG = IF and hence, IE = IF