II OBM

It is clear from the diagram that $$Q_n{Q}_{n+1}=P_n{P}_{n+1}\cos \angle B$$ $$R_n{R}_{n+1}=Q_n{Q}_{n+1}\cos \angle C$$ $$P_n{P}_{n+1}=R_{n-1}{R}_n\cos \angle A$$ Hence $P_n{P}_{n+1}=k^n{P}_0{P}_1$, where $|k|=|\cos A\cos B\cos C|<1$. If we take the direction $A$ to $B$ as positive, then the signed distance $P_n{P}_{n+1}$ may be positive or negative, but the series $1+|k|+|k{|}^2+|k{|}^3+\cdots$ converges to $\frac{1}{1-|k|}$, so $P_0{P}_1+P_1{P}_2+P_2{P}_3+\cdots$ is absolutely convergent and hence convergent. So the points $P_i$ converge to a point $P$ on the line $AB$.

Take any point $P^{\prime }$ on $AB$. Take $Q^{\prime }$ as the foot of the perpendicular from $P^{\prime }$ to $BC$. Now take $R^{\prime }$ as the intersection of the lines through $P^{\prime }$ perpendicular to $AB$ and through $Q^{\prime }$ perpendicular to $AC$. Now $P^{\prime }{Q}^{\prime }{R}^{\prime }$ is similar to the desired triangle $PQR$. Since $B,P^{\prime },P$ are collinear and $B,Q^{\prime },Q$ are collinear, it follows that $B,R^{\prime },R$ must be collinear. Thus extend $B{R}^{\prime }$ to meet $AC$ at

$R$. It is now straightforward to construct $Q$, then $P$.