$I$ is the incenter and $DI$ cuts $\ell$ at $E_0.$ Since $\angle BIC=\angle BDC=90^{\circ}+\frac{_1}{^2}\angle BAC,$ it follows that $B,I,C,D$ are concyclic $\Longrightarrow$ $\angle E_0IC=\angle ABC.$ Then $\angle ABC=\angle E_0AC$ $\Longrightarrow$ $A,I,C,E_0$ are concyclic. Since $\angle AID=\angle AIC,$ then chords $AC=AE_0$ are equal $\Longrightarrow$ $E \equiv E_0,$ as desired. Remark: Similarly, we show that $DF$ passes through the A-excenter of $\triangle ABC.$

Very easy. Call $ I$ is the incenter of $ \Delta ABC$. We have: $ \angle ADE=\pi-frac{\angle A+ \angle B}{2}=frac{\angle C}{2}=\\angle ACI=angle ADI=$ Hence we get the result.

In fact Where I first met this problem was in China's National High-school Math Competition(I don't konw the official English name of this contest, I just translated the words) in year 2005. It was in the second test.The second test contains 3 olympiad-type problems, each problem worths 50 points.120 minutes are allowed in the second test. I did this problem easily even though I was still a Junior student at that time.

djuro wrote: Let $ \triangle ABC$ with the incenter $ I$ and $ c>b$ . Denote $ D\in (AB)$ for which $ AD=b$ and $ E\in AA$ (the tangent in a point $ A$ to the circumcircle of $ \triangle ABC$) so that $ AE=b$ and the sideline $ AB$ doesn't separate $ C$ , $ E$ . Prove that the line $ I\in DE$ . Proof. $ X\in AI\cap DE$ and $ Y\in BC\cap AI\ \implies\ m(\angle BAE)=A+B$ , $ YD=YC=\frac {ab}{b+c}$ , $ m(\angle ADY)=C$ . Thus, $ DY\parallel AE\ \implies\ \frac {XA}{XY}=\frac {AE}{DY}=$ $ \frac {b}{\frac {ab}{b+c}}\ \implies\ \frac {XA}{XY}=\frac {b+c}{a}$ $ \stackrel{\mathrm{Van\ Aubel}}{\ \ \implies\ \ }X: =I$ . In conclusion, $ I\in DE$ .

Denote $\widehat{BAC}=2a,\widehat{ACB}=2c,\widehat{ABC}=2b$(after angle chasing) we have $\widehat{EAC}=2b$(using $\triangle{ACE},\triangle{ACD}$ are isosceles we have $\widehat{ADE}=90-a-b$,so$\widehat{EDC}=b,\widehat{ACI}=c,\widehat{ICD}=90+a-c$ and all we have to do is to aplt Ceva's reciprocal theorem in tre trigonometric form. The concluision follows only using thing like ${sin a sin b=-\frac{1}{2}(cos(a+b)-cos(a-b)}$

Let $\angle CAB=2\alpha$, $\angle ABC=2\beta$, $\angle ACB=2\gamma$. Suppose that $G\in BC$ such that $AG$ is bisector of $\angle A$. Let $I$ be the intersection point of $DE$ with $AG$. By hypothesis $\angle EAC=2\beta$, also the triangles $ADC$, $ADE$ and $ACE$ are isosceles, therefore making some calculations, $\angle AED=\gamma$, $\angle EDC=\alpha$ and $\angle IAC=\alpha$, hence AICE is cycle and $\angle ACI=\angle AEI=\angle AED=\gamma$, then $CI$ is bisector of $\angle C$ thus $I$ is the incenter. [geogebra]5e812a8a2647e28f7ba0108b9e11c654a4a76154[/geogebra]

linboll wrote: In fact Where I first met this problem was in China's National High-school Math Competition(I don't konw the official English name of this contest, I just translated the words) in year 2005. It was in the second test.The second test contains 3 olympiad-type problems, each problem worths 50 points.120 minutes are allowed in the second test. I did this problem easily even though I was still a Junior student at that time. yes,it is.and it can be second-killed by just some angle substitutions. by the way,the official name is China Second Round.

It can be also easily done with $S_{\triangle ADE}=S_{\triangle AID}+S_{\triangle AIE}$

Let $I$ be the intersection of the interior angle bisector with $\overline{DE}$. We have $\angle CEI = \angle CED = \frac12\angle CAD = \frac12\angle A = \angle CAI$, so $AICE$ is cyclic. Then, $\angle ICA = \angle IEA = \angle DEA = \frac12\angle DAF = \frac12\angle C$, so $\overline{IC}$ bisects $\angle C$. Hence, $I$ lies in two interior angle bisectors, so it is the incenter. $\blacksquare$