South-Afrika, 2011

Place the points $A$, $B$ and $C$ in the complex plane such that $P$ is at the origin. Then we have the following equations:
$$PP_1 \perp BC \implies \frac{B-C}{P_1} = \frac{\bar{C} - \bar{B}}{\bar{P}_1} \qquad (3)$$
$$PP_2 \perp AC \implies \frac{A-C}{P_2} = \frac{\bar{C} - \bar{A}}{\bar{P}_2} \qquad (4)$$
$$PP_3 \perp AB \implies \frac{B-A}{P_3} = \frac{\bar{A} - \bar{B}}{\bar{P}_3} \qquad (5)$$
$$\begin{align} BP_1 = BP_3 &\implies (B - P_1)(\bar{B} - \bar{P}_1) = (B - P_3)(\bar{B} - \bar{P}_3) \nonumber \\ &\implies B\bar{P}_3 + \bar{B}P_3 - P_3\bar{P}_3 = B\bar{P}_1 + \bar{B}P_1 - P_1\bar{P}_1 \tag{6} \end{align}$$
$$\begin{align} CP_1 = CP_2 &\implies (C - P_1)(\bar{C} - \bar{P}_1) = (C - P_2)(\bar{C} - \bar{P}_2) \nonumber \\ &\implies C\bar{P}_2 + \bar{C}P_2 - P_2\bar{P}_2 = C\bar{P}_1 + \bar{C}P_1 - P_1\bar{P}_1 \tag{7} \end{align}$$
Then,
$$\begin{align*} & AP_2^2 - AP_3^2 \\ &= (A - P_2)(\bar{A} - \bar{P}_2) - (A - P_3)(\bar{A} - \bar{P}_3) \\ &= A\bar{P}_3 + \bar{A}P_3 - A\bar{P}_2 - \bar{A}P_2 + P_2\bar{P}_2 - P_3\bar{P}_3 \\ &= B\bar{P}_3 + \bar{B}P_3 - C\bar{P}_2 - \bar{C}P_2 + P_2P_2 - 2P_3\bar{P}_3 \quad \text{(using (4) and (5))} \\ &= (B\bar{P}_3 + \bar{B}P_3 - P_3\bar{P}_3) - (C\bar{P}_2 + \bar{C}P_2 - P_2\bar{P}_2) \end{align*}$$
$$\begin{aligned} &= (B\bar{P}_1 + \bar{B}P_1 - P_1\bar{P}_1) - (C\bar{P}_1 + \bar{C}P_1 - P_2\bar{P}_1 \quad \text{(using (6) and (7))} \\ &= (B - C)\bar{P}_1 - (\bar{C} - \bar{B})P_1 \\ &= 0, \quad \text{(using (3))} \end{aligned}$$
as required.