It is confusing why they placed two easy geometry in a single test (P1 and P4). Maybe an inequality would have been better. Once you take out the midpoint of the arc, this is done easily.

Attachments:

LoloChen wrote: It is confusing why they placed two easy geometry in a single test (P1 and P4). Maybe an inequality would have been better. Once you take out the midpoint of the arc, this is done easily. D'KBF is cyclic, not D'CBF

This problem is not that easy if you attend the test. All four students in my school took 2 hours in average, and what confuses them is how to draw out an accurate graph.

However, if some knows not that much strange things, his thought may not diverse and this problem will even be easier for him.

Let $CI$ meet circumcircle at $T$. Claim $: FTIB$ is parallelogram. Proof $:$ Note that $TI || BF$ and $BF = CE = BT = TI$ (it's well-known that $TB = TI = TA$). Claim $: TAF$ and $BFK$ are congruent. Proof $:$ Note that $TF = BI = BK$ and $TA = TI = FB$ and $\angle FBK = \angle FBI + \angle IBK = \angle ITF + \angle IBK = \angle ITF + \angle ABC = \angle ITF + \angle ATB = \angle ATF$. Now Note that $FK = FA$ so we need to prove $FAD$ is isosceles. Note that $\angle TFA = \angle IBC = \frac{\angle B}{2}$ and $\angle FAB = \angle ABC$ so we need to prove $F,T,D$ are collinear. Lets assume not so $TF$ meets $BA$ at $S$. Claim $: SFBK$ is cyclic. Proof $:$ Note that $\angle FSB = \angle TFA = \angle BKF$ cause $TAF$ and $BFK$ are congruent. Now Note that $\angle SKB = \angle 180 - \angle SFB = \angle 180 - \frac{\angle B}{2} - \frac{\angle C}{2} = \angle 90 + \frac{\angle A}{2}$ and $\angle DKB = \angle CIB = \angle 90 + \frac{A}{2} \implies DKB = \angle SKB$ so $S$ and $D$ are same so $F,T,D$ are collinear as wanted so we're Done.