Trong tam giac ABC, CMR: $bc=\frac{l_{a}^{2}(b+c)^{2}}{(b+c)^{2}-a^2}$
CMR: $bc=\frac{l_{a}^{2}(b+c)^{2}}{(b+c)^{2}-a^2}$
#1
Đã gửi 24-06-2017 - 21:49
#2
Đã gửi 24-06-2017 - 22:18
*** Cannot compile formula: \definecolor{qqqqff}{rgb}{0.,0.,1.} \begin{tikzpicture}[line cap=round,line join=round,>=triangle 45,x=1.0cm,y=1.0cm] \clip(-2.2257713416241867,-2.9810135269838627) rectangle (10.961727291786438,7.377746728867376); \draw [line width=1.2pt] (1.9575741463157397,6.38171208888168)-- (-0.7,-2.1); \draw [line width=1.2pt] (-0.7,-2.1)-- (9.861212051653919,-2.1); \draw [line width=1.2pt] (9.861212051653919,-2.1)-- (1.9575741463157397,6.38171208888168); \draw [line width=1.2pt] (1.9575741463157397,6.38171208888168)-- (3.8831795199125017,-2.1); \begin{scriptsize} \draw [fill=qqqqff] (1.9575741463157397,6.38171208888168) circle (1.5pt); \draw[color=qqqqff] (1.778287911118314,6.889689755274385) node {$A$}; \draw [fill=qqqqff] (-0.7,-2.1) circle (1.5pt); \draw[color=qqqqff] (-0.9707676952422087,-2.3535117037928743) node {$B$}; \draw [fill=qqqqff] (9.861212051653919,-2.1) circle (1.5pt); \draw[color=qqqqff] (10.085216808599025,-2.2339875469945905) node {$C$}; \draw [fill=qqqqff] (3.8831795199125017,-2.1) circle (1.5pt); \draw[color=qqqqff] (3.909802040687705,-2.3535117037928743) node {$D$}; \end{scriptsize} \end{tikzpicture} *** 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*}
- Element hero Neos yêu thích
1 người đang xem chủ đề
0 thành viên, 1 khách, 0 thành viên ẩn danh