Let $\triangle ABC$ have incenter $I$. The line $CI$ intersects the circumcircle of $\triangle ABC$ for the second time at $L$, and $CI=2IL$. Points $M$ and $N$ lie on the segment $AB$, such that $\angle AIM =\angle BIN = 90^{\circ}$. Prove that $AB=2MN$.
Interesting Problem.
Let $IH$ be perpendicular to $AB$. First Lets have some angle chasing stuff. $\frac{\angle A}{2} = \angle HAI = \angle HIM,\frac{\angle B}{2} = \angle HBI = \angle HIN \implies \angle AIN = \frac{\angle C}{2} = \angle BIM$.
Claim : $ANI$ and $AIC$ are similar.
Proof : $\angle NAI = \angle IAC$ and $AIN = \angle ACI$.
Let $I'$ be reflection of $I$ across $N$ and $I_c$ be reflection of $C$ across $I$. Note that $CI=2IL$ so $IL = LI_c$. Now we have $L$ is center of $AIBI_c$.
Claim : $I'$ lies on $AIBI_c$.
Proof : we had $ANI$ and $AIC$ are similar and $I'N = NI , I_cI = IC$ so $AI'I$ and $AI_cC$ are similar so $\angle AI'I = \angle AI_cI$ so $I'$ lies on $AIBI_c$.
Claim : $AN = NH$.
Proof : we had $\angle HIN = \angle HBI$ so $NH.NB = NI^2 = NI.NI' = NA.NB \implies NH = NA$.
with same approach we have $MH = MB$ so $MN = \frac{AB}{2}$.
we're Done.