Theo công thức Heron, ta có:
\(S=\sqrt{p\left(p-a\right)\left(p-b\right)\left(p-c\right)}\) (với p là nửa chu vi)
\(=\dfrac{\sqrt{\left(a+b+c\right)\left(a+b-c\right)\left(a+c-b\right)\left(b+c-a\right)}}{4}\)
Theo giả thiết, ta có:
\(\left(a+b-c\right)\left(a+b+c\right)=4S\)
\(\Rightarrow\left(a+b-c\right)\left(a+b+c\right)=\sqrt{\left(a+b-c\right)\left(a+b+c\right)\left(a+c-b\right)\left(b+c-a\right)}\)
\(\Leftrightarrow\sqrt{\left(a+b-c\right)\left(a+b+c\right)}=\sqrt{\left(a+c-b\right)\left(b+c-a\right)}\)
\(\Leftrightarrow\left(a+b\right)^2-c^2=c^2-\left(a-b\right)^2\)
\(\Leftrightarrow a^2+2ab+b^2-c^2=c^2-a^2+2ab-b^2\)
\(\Leftrightarrow2a^2+2b^2=2c^2\)
\(\Leftrightarrow a^2+b^2=c^2\)
Suy ra tam giác đó là tam giác vuông (định lý Pytago đảo)