Chủ đề này có 1 trả lời

[happypolla]

Trong tam giac ABC, CMR: $(b-c).\cot \frac{A}{2}+(c-a).\cot \frac{B}{2}+(a-b). \cot \frac{C}{2}=0$

[huykinhcan99]

Áp dụng định lý $cosin$ cho tam giác $ABC$

\begin{align*} &\phantom{\iff~} \cos A = \dfrac{b^2+c^2-a^2}{2bc} \\ &\iff \dfrac{1-\cos A}{2} = \dfrac{a^2-\left(b-c\right)^2}{4bc} \\ &\iff \sin^2 \dfrac{A}{2} =\dfrac{\left(a-b+c\right)\left(a+b-c\right)}{4bc} \\ &\iff \sin^2 \dfrac{A}{2} =\dfrac{\left(p-b\right)\left(p-c\right)}{bc} \quad \text{(với $p=\dfrac{a+b+c}{2}$)} \end{align*}

 

Từ đó ta có

\begin{align*} \cot^2\dfrac{A}{2} &=\dfrac{1}{\sin^2 \dfrac{A}{2}}-1 \\ &= \dfrac{bc}{\left(p-b\right)\left(p-c\right)} - 1 \\ &= \dfrac{bc-\left(p-b\right)\left(p-c\right)}{\left(p-b\right)\left(p-c\right)} \\ & =\dfrac{p\left[\left(b+c\right)-p\right]}{\left(p-b\right)\left(p-c\right)}\\ & =\dfrac{p\left[\left(2p-a\right)-p\right]}{\left(p-b\right)\left(p-c\right)}\\ &=\dfrac{p\left(p-a\right)}{\left(p-b\right)\left(p-c\right)}\\ &=\dfrac{p^2\left(p-a\right)^2}{p\left(p-a\right)\left(p-b\right)\left(p-c\right)}\\ &=\dfrac{p^2\left(p-a\right)^2}{S^2} \end{align*}

 

Vậy

\[\cot \dfrac{A}{2}=\dfrac{p\left(p-a\right)}{S}\]

 

Tương tự

\begin{align*} \cot \dfrac{B}{2}&=\dfrac{p\left(p-b\right)}{S} \\ \cot \dfrac{C}{2}&=\dfrac{p\left(p-c\right)}{S}\end{align*}

 

Khi đó

\begin{align*} (b-c).\cot \frac{A}{2}+(c-a).\cot \frac{B}{2}+(a-b). \cot \frac{C}{2} &=\dfrac{p\left[\left(b-c\right)\left(p-a\right)+\left(c-a\right)\left(p-b\right)+\left(a-b\right)\left(p-c\right)\right]}{S} \\ &=\dfrac{p\left\{\left(b-c\right)\left(p-a\right)+\left(c-a\right)\left(p-b\right)-\left[\left(b-c\right)+\left(c-a\right)\right]\left(p-c\right)\right\}}{S} \\ &=\dfrac{p\left\{\left(b-c\right)\left[\left(p-a\right)-\left(p-c\right)\right]+\left(c-a\right)\left[\left(p-b\right)-\left(p-c\right)\right]\right\}}{S} \\ &=\dfrac{p\left[\left(b-c\right)\left(c-a\right)+\left(c-a\right)\left(c-b\right)\right]}{S}\\ &=0 \end{align*}