Problem 1930

(a) Denote $(I)$ as the incircle of triangle $ABC$. Let $D', F'$ be the tangent points of $(I)$ with $BC, BA$. $DW$ is the diameter of $(J)$. We have
$$\frac{AI}{AJ} = \frac{IF'}{JF} = \frac{ID'}{JW}$$
hence $A, D', W$ are collinear. Denote $D_1$ as the intersection of $ID'$ and $AD$. We have $JD = JW$, so $ID' = ID_1$. Moreover, $BD' = CD$ then $LD = LD'$, therefore $IL \parallel DD_1$.

Let $X$ be the midpoint of $D'W$. It is easy to see that $DLXJ$ is a rectangle. Thus $\angle XHD = 90^\circ$, and $XH \parallel JB$. Then we have $XH \perp BI$.

Construct a rectangle $CDJU$. We have $JU \parallel CD$ and $JU = CD$, hence $JU \parallel BD'$ and $JU = BD'$. Then $BD'UJ$ is a parallelogram. We conclude that $D'U \parallel XH$.

Let $Y, V, Z$ be the intersection points of $XH$ with $DJ, UW, UC$, respectively. Notice that $V$ is the midpoint of $UW$, and combined with $UZ \parallel YW$, then $V$ is the midpoint of $YZ$.

The cyclic quadrilateral $CEUJ$ has $CE = CD = UJ$, so $UE \parallel CJ$ and $UE \perp DE$. Moreover, $EW \perp ED$ then $W, U$ and $E$ are collinear.

Triangles $BIC$ and $YVW$ have
$$\angle YWV = \angle CED = \angle ICB$$
and
$$\angle WYV = \angle XYJ = \angle BJD = \angle IBC$$
then $\triangle BIC \sim \triangle YVW$ (a-a). But $V, L$ are midpoints of segments $YZ, BC$ respectively, therefore $\triangle BIL \sim \triangle YZW$ (s-a-s). Hence $\angle YWZ = \angle BLI = \angle BDA = 90^\circ - \angle WDU'$. Let $U'$ be the intersection of $WZ$ and $AD$, then $\angle DU'W = 90^\circ$. Moreover, we have $\angle DU'W = \angle DU'Z = 90^\circ$. Thus $U'$ lies on $(CDH)$ and $(J)$.

Analogously, $(BDK)$ passes through $U'$. This completes the proof of this part.

(b) Let $S', R$ be the second intersection points of $AI, SI$ with $(O)$. We have $S'I^2 = S'L \cdot S'S$ then $\angle S'IL = \angle S'SI = \angle SAR$. But we have $IL \parallel AD$ so $\angle S'IL = \angle S'AD$. Then we conclude that $\angle S'AR = \angle S'AD$ and $\angle BAR = \angle CAD$.

Construct $DH' \perp JG$ ($H' \in JG$). It is easy to see that $GH' \cdot GJ = GD^2 = GE \cdot GF$ then $EFJH'$ is a cyclic quadrilateral. We also have $AEJF$ is cyclic, so $A, E, F, J$ are $H'$ concyclic. Therefore $\angle AH'J = 90^\circ$ and $A, D, H'$ are collinear.

On the other hand, we have $\angle PAJ = \angle QAJ = 90^\circ + \frac{1}{2}\angle BAC = \angle BIC$ and $\angle IBC = \angle IJC$, $\angle ICB = \angle IJB$ then $\triangle IBC \sim \triangle APJ \sim \triangle AJQ$ (a-a).

Denote $P', Q'$ as midpoints of $JP, JQ$ then $\triangle IBL \sim \triangle APP' \sim \triangle AJQ'$ and $\triangle ICL \sim \triangle AQQ'$. Therefore $A, P', J, Q'$ are concyclic. Let $L'$ be the intersection of $LI$ and $PQ$. With $\angle BLI = \angle JAQ' = \angle JP'Q' = \angle BPQ$, we have $P, B, I, L'$ are concyclic. We also have $\angle PBI = 90^\circ$ therefore $IL \perp PQ$. Moreover, $IL \parallel AD$ and $AD \perp MN$ then $MN \parallel PQ$.

Consider $A' \in AT$ such that $MA' \parallel AT$. Notice that
$$\frac{TA'}{TA} = \frac{TM}{TP} = \frac{TN}{TQ},$$
so $NA' \parallel AQ$, then $\angle AMA' = \angle ANA' = 90^\circ$.

Thus $\angle BAR = \angle BAT = 90^\circ - \angle AA'M = 90^\circ - \angle ANM = \angle CAD$. Then $A, R, T$ are collinear, the proof is complete.