Problem 4

Use directed angles modulo $180^\circ$. By the two cyclic quadrilaterals from Step 3, $\angle CTP=\angle CYP$ and $\angle PTB=\angle PXB$. Because $B,X,Y,C$ are in this order on one line, reversing the first ray in each of these two angles does not change its directed value, so $\angle CYP=\angle XYP$ and $\angle PXB=\angle YXP$ as directed angles. Therefore $\angle CTB=\angle CTP+\angle PTB=\angle XYP+\angle YXP=\angle XPY$ modulo $180^\circ$, using the directed angle sum in triangle $PXY$. We now choose the ordinary representatives. The altitude from $A$ to $BC$ is $2\Delta/a$ and the inradius is $r=\Delta/s$, so $2\Delta/a>2r$ because $s>a$; hence the reflection $T$ of $A$ over $I$ lies on the side of $BC$ opposite $A$. The second circumcircle intersection $P$ also lies on that side, since the ray from $A$ through the interior point $I$ crosses $BC$ before reaching the circumcircle again. With $B<X<Y<C$, this configuration makes the two ordinary angles represented by the directed equality supplementary (the angle at $T$ opens toward the whole segment $BC$, while the angle at $P$ subtends the inner segment $XY$). Thus $\angle BTC+\angle XPY=180^\circ$. Finally Step 1 gives $\angle KIL=\angle BTC$, and $\angle XPY=\angle YPX$, so the required sum is $180^\circ$.