Đặt \(p=\frac{a+b+c}{2}\)
=>a+b+c=2p
=>a+b-c=2p-2c
=>a+b-c=2(p-c)
Ta có: a+b+c=2p
=>a-b+c=2p-2b
=>a-b+c=2(p-b)
Ta có: a+b+c=2p
=>b+c-a=2p-2a
=>b+c-a=2(p-a)
\(S=\frac14\left(a+b-c\right)\left(a-b+c\right)\)
=>\(S=\frac14\cdot2\left(p-c\right)\cdot2\cdot\left(p-b\right)=\left(p-c\right)\left(p-b\right)\) (1)
Vì S là diện tích tam giác ABC và p là nửa chu vi của tam giác ABC
nên \(S_{ABC}=\sqrt{p\left(p-a\right)\left(p-b\right)\left(p-c\right)}\)
=>\(\sqrt{p\left(p-a\right)\left(p-c\right)\left(p-b\right)}=\left(p-c\right)\left(p-b\right)\)
=>\(p\left(p-a\right)\left(p-c\right)\left(p-b\right)=\left(p-c\right)^2\left(p-b\right)^2\)
=>p(p-a)=(p-c)(p-b)
=>\(2p\cdot2\left(p-a\right)=2\left(p-c\right)\cdot2\cdot\left(p-b\right)\)
=>(a+b+c)(b+c-a)=(a+b-c)(a-b+c)
=>\(\left(b+c\right)^2-a^2=a^2-\left(b-c\right)^2\)
=>\(\left(b+c\right)^2+\left(b-c\right)^2=2a^2\)
=>\(2a^2=b^2+2bc+c^2+b^2-2bc+c^2=2b^2+2c^2\)
=>\(a^2=b^2+c^2\)
=>ΔABC vuông tại A
=>\(\hat{BAC}=90^0\)
