Chủ đề này có 1 trả lời

[happypolla]

Trong tam giac ABC, CMR: $bc=\frac{l_{a}^{2}(b+c)^{2}}{(b+c)^{2}-a^2}$

[huykinhcan99]

*** Cannot compile formula:



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*** Error message:
Error: Nothing to show, formula is empty

 

Đầu tiên, áp dụng công thức tính diện tích tam giác ta có

\begin{align*} &\phantom{\iff~} \left\{ \begin{array}{l} S_{ABC} = \dfrac{1}{2} bc \sin A = bc \sin \dfrac{A}{2} \cos \dfrac{A}{2}\\ S_{ACD} = \dfrac{1}{2} b l_a \sin \dfrac{A}{2}\\ S_{ABD} = \dfrac{1}{2} c l_a \sin \dfrac{A}{2}   \end{array} \right. \\ &\implies bc \sin \dfrac{A}{2} \cos\dfrac{A}{2} = \dfrac{1}{2} b l_a \sin \dfrac{A}{2}+ \dfrac{1}{2} c l_a \sin \dfrac{A}{2} \\ &\iff bc\cos \dfrac{A}{2} = \dfrac{l_a\left(b+c\right)}{2} \quad \text{(với chú ý rằng $A>0\implies \sin\dfrac{A}{2}>0$)} \\ &\implies b^2c^2 \cos^2\dfrac{A}{2} =\dfrac{l^2_a\left(b+c\right)^2}{4} \\ &\iff \dfrac{b^2c^2\left(1+\cos A\right)}{2}=\dfrac{l^2_a\left(b+c\right)^2}{4}\end{align*}

 

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

\[\cos A = \dfrac{b^2+c^2-a^2}{2bc}\]

 

Từ đó ta thu được

\begin{align*} &\phantom{|iff~} \dfrac{b^2c^2\left(1+ \dfrac{b^2+c^2-a^2}{2bc}\right)}{2}=\dfrac{l^2_a\left(b+c\right)^2}{4} \\ & \iff \dfrac{bc\left[2bc+\left(b^2+c^2-a^2\right)\right]}{4} =\dfrac{l^2_a\left(b+c\right)^2}{4}  \\ &\iff bc\left[\left(b+c\right)^2-a^2\right]=l^2_a\left(b+c\right)^2 \\ &\iff bc=\dfrac{l^2_a\left(b+c\right)^2}{\left(b+c\right)^2-a^2} \tag{$\blacksquare$}\end{align*}