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a)

b) \(\hept{\begin{cases}AM^2=AH^2+HM^2=\left(AC^2-HC^2\right)+\left(MC-HC\right)^2=AC^2+MC^2-2MC.HC=AC^2+\frac{BC^2}{4}-BC.HC\\AM^2=AH^2+HM^2=\left(AB^2-BH^2\right)+\left(BH-BM\right)^2=AB^2+BM^2-2BH.BM=AB^2+\frac{BC^2}{4}-BC.BH\end{cases}}\)

\(\Rightarrow2AM^2=AB^2+AC^2+\frac{BC^2}{2}-BC\left(HC+BH\right)=AB^2+AC^2+\frac{BC^2}{2}-BC^2=AB^2+AC^2-\frac{BC^2}{2}\)

\(CMTT\Rightarrow\hept{\begin{cases}4BM^2=2AB^2+2BC^2-AC^2\\4CN^2=2AC^2+2BC^2-AB^2\end{cases}\left(4\right)}\)

\(BM\perp CN\Leftrightarrow BG^2+CG^2=BC^2\Leftrightarrow\left(\frac{2}{3}BN\right)^2+\left(\frac{2}{3}CN\right)^2=BC^2\Leftrightarrow4BN^2+4CN^2=9BC^2\left(5\right) \)

\(Từ\left(4\right),\left(5\right)\Rightarrow\left(2AB^2+2BC^2-AC^2\right)+\left(2AC^2+2BC^2-AB^2\right)=9BC^2\Leftrightarrow5BC^2=AB^2+AC^2\left(đpcm\right)\)

Xét ΔABC vuông tại A có

\(cotB=\dfrac{BA}{AC};cotC=\dfrac{AC}{AB}\)

\(cotB+cotC=\dfrac{BA}{AC}+\dfrac{AC}{AB}\)

\(=\dfrac{AB^2+AC^2}{AB\cdot AC}=\dfrac{BC^2}{AB\cdot AC}\)

\(=\dfrac{BC}{AB\cdot AC}\cdot BC=\dfrac{BC}{AH}\)

Kẻ đg cao AH, trung tuyến AD, trọng tâm G

Tg AHD vuông tại H nên \(AH\le AD\Rightarrow\dfrac{BC}{AH}\ge\dfrac{BC}{AD}\left(4\right)\)

Ta có \(\cot\widehat{B}+\cot\widehat{C}=\dfrac{BH}{AH}+\dfrac{CH}{AH}=\dfrac{BC}{AH}\ge\dfrac{BC}{AD}\left(1\right)\)

Mà BM vuông góc CN nên GD là trung tuyến ứng vs ch BC

\(\Rightarrow BC=2GD\left(2\right)\)

Mà G là trọng tâm nên \(3GD=AD\left(3\right)\)

 \(\left(1\right)\left(2\right)\left(3\right)\left(4\right)\Rightarrow\cot\widehat{B}+\cot\widehat{C}\ge\dfrac{BC}{AD}=\dfrac{2GD}{3GD}=\dfrac{2}{3}\)

C