NMO Selection Tests for the Junior Balkan Mathematical Olympiad

We shall prove that $\triangle PAB \sim \triangle PA'B'$, and similarly $\triangle PBC \sim \triangle PB'C'$ and $\triangle PCA \sim \triangle PC'A'$; obviously, these three triangle similarities are enough in order to prove that $\triangle ABC \sim \triangle A'B'C'$, either from angle equalities or side ratios.

Obviously, we have $\triangle PAE \sim \triangle PBD$, so $\frac{AE}{PE} = \frac{BD}{PD}$. For symmetry reasons, we have $AE = A'E$ and $BD = B'D$, so $\frac{A'E}{PE} = \frac{B'D}{PD}$.

Also,
$$\begin{align*} \angle A'EP &= |\angle PEF - \angle A'EF| = |\angle BEF - \angle AEF| = |\angle BDF - \angle ADF| \\ &= |\angle B'DF - \angle PDF| = \angle B'DP. \end{align*}$$
It follows that $\triangle PA'E \sim \triangle PB'D$, whence
$$\frac{PA'}{PE} = \frac{PB'}{PD} \text{ and } \angle A'PE \equiv \angle B'PD,$$
so $\angle A'PB' \equiv \angle EPD$. Therefore, $\triangle A'PB' \sim \triangle EPD$, and, since $\triangle EPD \sim \triangle APB$, we get $\triangle PAB \sim \triangle PA'B'$.