nice problem. i will show the one part. the other is similar. wlog $ \angle B > \angle C$ it is enough to show that, $ \angle ICJ = \angle BIJ \leftrightarrow \angle C = 2\angle IBP \leftrightarrow \angle ABP = \frac {\angle(B - C)}{2}$ but if we take $ PA'$ the diametre of the circumcircle, and $ AH_1$ the perpendicular from $ A$ to $ BC$ from known lemma we get $ \frac {\angle(B - C)}{2} = \angle H_1AA' = \angle AA'P = \angle ABP$ QED..

let Q be the intersection of AI with (O) (the circumference of ABC) then Q,P,O collinear and PQ is perpendicular with BC which means JI is the bisector of <BJC let O1 O2 be the center of the circumference of IJB,JIC.We have <BIO1=(180-<BO1I)/2=(180-<BJC)/2=A/2.<O1IO2=180-<IO1O2-<IO2O1=180-<IBJ-<ICJ=180-(360-<BIC-<BJC)=90-A/2 (notice <BIC=90+A/2) We are done

<BJP=<BCP=(180-<BAC)/2= (<ABC+<ACB)/2 , <CJP=<CAP=(180-<BAC)/2= (<ABC+<ACB)/2 Then <BJP=<CJP. <PBI=<CBP- <CBI= (<ABC+<ACB)/2- <ABC/2 =<ACB/2 <BIJ= <PBI+<BPI= <ACB/2 + <BCJ= < ICB then <BIJ=<ICB↔ the circumcircles ∆ JIC is tangent to BI and <JBI=<JIC ↔ the circumcircles ∆JIB is tangent to CI

let $ K$ be the intersection of $ CI$ and the circumcircle of $ ABC$... it's easy to note that $ P$ is the midpoint of the arc $ BAC$, and since arc $ BK=\angle C$, then, arc $ PK=\angle B$... this implies that $ \angle CIJ = \textrm{arc} \ CJ + \textrm{arc} \ PK = \angle CBJ + \angle \frac {B}{2} = \angle CBJ + \angle IBC = \angle IBJ$, and we're done for the sake of completeness i think it should be proved that $ J$ lies in (small) arc $ BC$... this follows from the fact that $ P$ lies between the two midpoints of arcs $ AB$ and $ AC$...

let $\angle API=\alpha$ . Then , $\angle ACJ= \alpha$ . Then , $\angle ICJ= \alpha-\frac{C}{2}$ In cyclic quadrilateral $APJB$ , $\angle BJP=\frac{\pi-A}{2}$ and $\angle ABJ=\pi-\alpha$ . hence , $\angle BIJ=\alpha-\frac{C}{2}$ .So , The circumcircle of $\Delta IJC$ has $IB$ as a tangent

Dear Mathlinkers, J is the point of contact of the A-mixtilinear incircle of ABC... see http://perso.orange.fr/jl.ayme vol. 4 A new mixtilinear incircle adventure I, p. 81. Sincerely Jean-Louis