Let $S$ be such that $\triangle ABC$ and $\triangle BCS$ are in a different half planes with respect to $BC$, and $\triangle BCS$ is isosceles and has right angle $CBS$. Then segments $BS$ and $PT$ are equal and parallel ($TP = BC = BS$, $PS \parallel AB$), hence $TPSB$ is a parallelogram (Fig. 21). Thus, $PS \parallel AB$, and also $PS = TB$, hence $AT_1 = PS$. Thus $AT_1SP$ is a parallelogram. Thus, $AP \parallel ST_1$ and $AP = ST_1$. Therefore, $P_1C \parallel T_1S$ and $P_1C = T_1S$ is a parallelogram. Hence $P_1T_1 \parallel CS$ and $\angle P_1T_1A = 180^\circ - \angle P_1T_1S - \angle ST_1B = \angle CST_1 - \angle ST_1B$.

Fig. 21
Finally,
$$\begin{aligned} \angle CBA - \angle P_1T_1A &= \angle CBA - \angle CST_1 + \angle ST_1B = \\ &= 90^\circ - \angle T_1SB - \angle CST_1 = 90^\circ - \angle CSB = 45^\circ. \end{aligned}$$

Let $PT$ and $BC$ intersect at $K$, $M_c$ is a midpoint of $CP$, $M_b$ is a midpoint of $BT$ (Fig. 22). Thus, medians in right triangles $CKP$ and $BKT$ equal half of hypotenuses. Therefore:
$$\angle M_c KM_b = \angle TKM_b - \angle PKM_c = \angle M_b TK - \angle M_c PK = \angle CAB.$$
Also, $\frac{AR}{AT_1} = \frac{CP}{BT} = \frac{KM_c}{KM_b}$, hence $\triangle AP_1T_1 \sim \triangle KM_cM_b$. We can show that $\angle CBA - \angle P_1T_1A = 45^\circ$. Also, $\angle CBA - \angle P_1T_1A = \angle M_bKB - \angle M_cM_bK$, that is an angle between $M_cM_b$ and $BC$.
Thus, it suffices to show that $M_cM_b$ is parallel to bisector of $\angle TKB$.

Let $X$ is a midpoint of $PB$, $Y$ is a midpoint of $TC$ (Fig. 23) where $M_c M_b$ is a bisector of $\angle XM_c Y$. $XM_c \parallel KB$ and $YM_c \parallel KT$ finish the proof.