Referring to Figure 3 we have,
BX/XC=(ABX)/(AXC)=(PBX)/(PXC)=(ABX)-(PBX)/(AXC)-(PXC)= (APB)/(CAP).
Similarly, CY/YA=(BCP)/(ABP), AZ/ZB=(CAP)/(BCP).
Now, multiplying these together, we have
BD/DC*CE/EA*AF/FB=(ABK)/(CAK)*(BCK)/(ABK)*(CAK)/(BCK)=1
This completes the first part of the proof. The second part of
the theorem is proved in the same way as it was in the first proof.