Well-known (this is called Iran lemma). Unrelated question: can you post the problems from the senior division of Korea MO as well, if you have them?

We can simply solve this problem by angle chasing. Let the triangle $ABC$ meets the segments $AC$ at $F$.Let $P$ be the intersection of the line $AI$ and $DE$ It's easy to prove that $\triangle ADP\cong\triangle AFP$ $\angle APF=\angle APD=\angle BDP-\frac 12\angle BAP=90^o-\frac 12\angle B-\frac 12\angle A=\frac 12\angle C=\angle ACI$ So $I,P,F,C$ is concylic. $\because IF\bot AC$,therefore $IP\bot AC$. $\therefore P$ is on the circle whose diameter is $AC$.

Let $AI$ and $DE$ intersect at $P$. Let $\angle BAC=\alpha$, $\angle CBA=\beta$, $\angle ACB=\gamma.$ Claim: $ICPE$ is cyclic. Note that $\angle ICE=\gamma/2$. Note that $BEID$ is cyclic. Additionally, $\angle IBE=\beta/2$, so $\angle IDE=\beta/2$. From $\triangle ADP$, we have $$\angle IPE=180-\alpha/2-90-\beta/2=\gamma/2,$$so $ICPE$ is cyclic. Finally, from $ICPE$ cyclic, $\angle IEC=90$ implies that $\angle IPC=90$, so we are done.