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

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