China Mathematical Competition (Complementary Test)

As shown in Fig. 1.2, we connect $AD$, $BC$, $CF$, $DE$. Since $AE = EF = FB$, we have
$$\frac{BC \cdot \sin \angle BCE}{AC \cdot \sin \angle ACE} = \frac{\text{distance between } B \text{ and } CP}{\text{distance between } A \text{ and } CP} = \frac{BE}{AE} = 2. \qquad \textcircled{1}$$
In the same way,
$$\frac{AD \cdot \sin \angle ADF}{BD \cdot \sin \angle BDF} = \frac{\text{distance between } A \text{ and } PD}{\text{distance between } B \text{ and } PD} = \frac{AF}{BF} = 2. \qquad \textcircled{2}$$
On the other hand, since
$$\angle BCE = \angle BCP = \angle BDP = \angle BDF,$$
$$\angle ACE = \angle ACP = \angle ADP = \angle ADF,$$
multiplying ① by ②, we have $\frac{BC \cdot AD}{AC \cdot BD} = 4$, or
$$BC \cdot AD = 4AC \cdot BD. \qquad \textcircled{3}$$
By Ptolemy's Theorem, we have
$$AD \cdot BC = AC \cdot BD + AB \cdot CD. \qquad \textcircled{4}$$
Combining ③ and ④, we get $AB \cdot CD = 3AC \cdot BD$, and that is
$$EF \cdot CD = AC \cdot BD.$$
The proof is complete.

Fig. 1.2