58th Ukrainian National Mathematical Olympiad

Let $S$ be such that $△ABC$ and $△BCS$ are in a different half planes with respect to $BC$, and $△BCS$ 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 $A{T}_1=PS$. Thus $A{T}_{1}SP$ is a parallelogram. Thus, $AP\parallel S{T}_1$ and $AP=S{T}_1$. Therefore, $P_{1}C\parallel T_{1}S$ and $P_{1}C=T_{1}S$ is a parallelogram. Hence $P_1{T}_1\parallel CS$ and $\angle P_1{T}_{1}A=180^{\circ }-\angle P_1{T}_{1}S-\angle S{T}_{1}B=\angle CS{T}_1-\angle S{T}_{1}B$.

Fig. 21 Finally, $$\begin{array}{rl}\angle CBA-\angle P_1{T}_{1}A& =\angle CBA-\angle CS{T}_1+\angle S{T}_{1}B=\\ & =90^{\circ }-\angle T_{1}SB-\angle CS{T}_1=90^{\circ }-\angle CSB=45^{\circ }.\end{array}$$

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Alternative solution.

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: $$\angle M_{c}K{M}_b=\angle TK{M}_b-\angle PK{M}_c=\angle M_{b}TK-\angle M_{c}PK=\angle CAB.$$ Also, $\frac{AR}{A{T}_1}=\frac{CP}{BT}=\frac{K{M}_c}{K{M}_b}$, hence $△A{P}_1{T}_1\sim △K{M}_c{M}_b$. We can show that $\angle CBA-\angle P_1{T}_{1}A=45^{\circ }$. Also, $\angle CBA-\angle P_1{T}_{1}A=\angle M_{b}KB-\angle M_c{M}_{b}K$, that is an angle between $M_c{M}_b$ and $BC$. Thus, it suffices to show that $M_c{M}_b$ 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 $\angle X{M}_{c}Y$. $X{M}_c\parallel KB$ and $Y{M}_c\parallel KT$ finish the proof.