Giải thích các bước giải:
h.Gọi $AD\perp BC, BE\perp AC, CF\perp AB$
$\to BE.AC=CF.AB=AD.BC=2S_{ABC}=2S$
$\to\cot A=\dfrac{AE}{BE}=\dfrac{AF}{CF}$
$\cot C=\dfrac{CE}{BE}=\dfrac{CD}{CA}$
$\cot B=\dfrac{BD}{AD}=\dfrac{BF}{CF}$
$\to\cot A+\cot C=\dfrac{AE}{BE}+\dfrac{CE}{BE}=\dfrac{AE+CE}{BE}=\dfrac{AC}{BE}=\dfrac{AC^2}{BE.AC}=\dfrac{b^2}{2S}$
Tương tự $\cot B+\cot C=\dfrac{a^2}{2S}$
$\cot A+\cot B=\dfrac{c^2}{2S}$
$\to (\cot B+\cot C)+(\cot A+\cot B)=\dfrac{a^2+c^2}{2S}$
$\to 2\cot B+(\cot C+\cot A)=\dfrac{a^2+c^2}{2S}$
$\to 2(\cot C+\cot A)=\dfrac{a^2+c^2}{2S}$
$\to \dfrac{2b^2}{2S}=\dfrac{a^2+c^2}{2S}$
$\to 2b^2=a^2+c^2$
