Đặt BC = a , AC = b , AB = c . Ta có :
\(BD=\frac{a+c-d}{2}\)
\(DC=\frac{a+b-c}{2}\)
Do đó , ta giả sử \(\left(b\ge c\right)\)
\(BD.DC=\frac{a+c-b}{2}.\frac{a+b-c}{2}\)
\(=\frac{a-\left(b-c\right)}{2}.\frac{a+\left(b-c\right)}{2}\)
\(=\frac{a^2-\left(b-c\right)^2}{4}\)
\(=\frac{a^2-b^2+2bc-c^2}{4}\)
\(=\frac{a^2-\left(b^2+c^2\right)+2bc}{4}\)
Do \(a^2=b^2+c^2\)nên \(BD.DC=\frac{2bc}{3}=\frac{bc}{2}=S_{ABC}\)