China Mathematical Competition (Complementary Test)

As shown in Fig. 1.2, we connect $AD$, $BC$, $CF$, $DE$. Since $AE=EF=FB$, we have $$\frac{BC\cdot \sin \angle BCE}{AC\cdot \sin \angle ACE}=\frac{\text{distance between }B\text{ and }CP}{\text{distance between }A\text{ and }CP}=\frac{BE}{AE}=2.\stackrel{◯}{1}$$ In the same way, $$\frac{AD\cdot \sin \angle ADF}{BD\cdot \sin \angle BDF}=\frac{\text{distance between }A\text{ and }PD}{\text{distance between }B\text{ and }PD}=\frac{AF}{BF}=2.\stackrel{◯}{2}$$ 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 $$BC\cdot AD=4AC\cdot BD.\stackrel{◯}{3}$$ By Ptolemy's Theorem, we have $$AD\cdot BC=AC\cdot BD+AB\cdot CD.\stackrel{◯}{4}$$ 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