trên AC lấy 1 điểm M sao cho : $\widehat{ABM}=\widehat{CBD}$
Ta có:$\widehat{ABM}+\widehat{CBM}=\widehat{ABC}=\widehat{ABD}+\widehat{CBD}=\widehat{ABD}+\widehat{ABM}$
$\Rightarrow \widehat{ABD}=\widehat{CBM}$$\triangle _{ABD}\sim \triangle _{MBC}$ (g,g)
$\Rightarrow \frac{AD}{MC}=\frac{BD}{MC}$
$ \Rightarrow AD.BC = BD . MC (1)$
$\triangle _{ABM}\sim \triangle _{DBC}$(g,g)
$\Rightarrow \frac{AB}{DB}=\frac{AM}{DC}$
$\Rightarrow AB . DC = DB.AM (2)$
TA LẤY (1) + (2) THEO VẾ TA ĐƯỢC : $AD.BC + AB.DC = DB.MC + DB.AM = DB.(MC +AM)$
$\Rightarrow AD.BC + AB.DC = DB.(MC +AM) = DB.AC$ (điều phải chứng minh )