Consider the cyclic quadrilateral ABCD. Construct the point E on AC such that [math]\angle ABE\cong\angle DBC[/math]. Then, we can identify the similar triangles: [math]\triangle ABE\sim\triangle DBC[/math] and [math]\triangle BCE\sim\triangle BDA[/math].[br][br]From these similarities we can show the equal ratios: [br][math]\frac{AE}{AB}=\frac{CD}{BD}[/math],[br][math]\frac{EC}{BC}=\frac{DA}{BD}[/math].[br]That is, [math]AE\cdot BD=AB\cdot CD[/math] and [math]BD\cdot EC=BC\cdot DA[/math]. [br][br]By adding we can show that [math]AE\cdot BD+BD\cdot EC=AB\cdot CD+BC\cdot DA[/math]. [br]However, notice that [math]AE\cdot BD+BD\cdot EC=BD\cdot\left(AE+EC\right)=BD\cdot AC[/math]. [br][br]Thus, [math]AC\cdot BD=AB\cdot CD+BC\cdot DA[/math].