IRL ABooklet

Because $ABDC$ is cyclic,
$$\angle TAL = \angle BAD = \angle BCD = \angle DCL = \angle TCL$$
hence quadrilateral ACLT is cyclic. This implies
$$\angle OLP = \angle ALT = \angle ACT = 90^{\circ} \quad \text{and}$$
$$\angle CLP = \angle CLT = 180^{\circ} - \angle A = 180^{\circ} - \angle COM = 180^{\circ} - \angle COP$$
(by the central angle theorem), and so OCLP is also a cyclic quadrilateral.

Hence $\angle OCP = \angle OLP = 90^\circ$, i.e. $PC$ is tangent to circumcircle (ABC). As $OP$ is the perpendicular bisector of $BC$ ($M$ is the midpoint of the chord $BC$ of circle centre $O$), $\angle BOP = \angle COP$ and by SAS, $\triangle COP \equiv \triangle BOP$. This shows that $PB$ is also tangent to circumcircle (ABC).

As $\angle ACD = 90^\circ$, $AD$ is a diameter of circle (ABC) and so $\angle ABD = 90^\circ$ as well. Let $BD$ (extended) intersect line $AC$ (extended) at $S$. Then $D$, the intersection point of $BS$ and $CT$, is the orthocentre of $\triangle AST$. Like in Solution 1, from $\angle BAD = \angle BCD = \angle DCL$ we see that $ACLT$ is cyclic and find $\angle ALT = \angle ACT = 90^\circ$. It follows that $L$ lies on $ST$.

Because $AD$ is a diameter of circle (ABC), $O$ is the midpoint of $AD$. Therefore, the points $B$, $O$, $C$, $L$ are all on the 9-point circle of $\triangle AST$. Because $OP$ is the perpendicular bisector of the chord $BC$ of this circle, it contains the centre of it. Since $\angle OLP = 90^\circ$, we now see that $P$ is on this 9-point circle as well and $OP$ is a diameter of it. This shows that $\angle OCP = \angle OBP = 90^\circ$, i.e. $PB$ and $PC$ are tangents to (ABC).

As in the other solutions, we first establish that the quadrilateral ACLT is cyclic and $\angle TLA = \angle TCA = 90^\circ$. Let AL intersect BC at N.

Since CD is the internal angle bisector of $\angle BCL$ and $\angle ACD = 90^\circ$, then $CA$ is the external angle bisector of $\angle BCL$ and thus $L$, $D$, $N$, $A$ form a harmonic range (i.e. cross-ratio $[L, N; D, A] = -1$; or $L$, $N$ are harmonic conjugates with respect to $A$ and $D$), hence $L$ is on the polar line of $N$ with respect to the circumcircle (ABC). Moreover, $OL \perp LT$ so $LT$ is the polar line of $N$. Since $P$ is on $LT$, $N$ is on the polar line of $P$, which is perpendicular to $OP$, hence must be the line $BC$. It follows that $PB$ and $PC$ are both tangent to circumcircle (ABC).