+118687 If E is the middle point of the side CA of the triangle ABC and if R is the area of the triangle, prove that:
Cot AEB = (BC2 - BA2) / 4R
Let
x=AB
y=CB
w=EB
z=AE=EC
angle AEB= theta
angle CEB= 180-theta
\(R=0.5zwsin\theta+0.5zwsin(180-\theta)\\ R=0.5zwsin\theta+0.5zwsin(\theta)\\ R=zwsin\theta\\ 4R=4zwsin\theta\\ \)
\(y^2=w^2+z^2-2wzcos(180-\theta)\\ y^2=w^2+z^2+2wzcos(\theta)\\ x^2=w^2+z^2-2wzcos(\theta)\\ subtracting\\ y^2-x^2=4wzcos(\theta)\\ BC^2-BA^2=4wzcos(\theta)\\ \)
\(\therefore\\ RHS =\frac{BC^2-BA^2}{4R}\\ RHS =\frac{4wzcos(\theta)}{4wzsin(\theta)}\\ RHS =cot(\theta)\\ RHS=LHS \qquad QED\)
+26393 If E is the middle point of the side CA of the triangle ABC and
if R is the area of the triangle, prove that:
\(\cot(AEB) = \dfrac{ BC^2 - BA^2 } { 4R } \)
\(\text{Let $\angle AEB= \epsilon$} \\ \text{Let $AE=EC$} \)
\(\begin{array}{|rcll|} \hline BC^2 &=& AE^2 + BE^2 - 2\cdot AE\cdot BE \cdot \cos(180^\circ-\epsilon ) \\ &=& AE^2 + BE^2 + 2\cdot AE\cdot BE \cdot \cos(\epsilon ) \\\\ BA^2 &=& EC^2 + BE^2 - 2\cdot EC\cdot BE \cdot \cos(\epsilon ) \\\\ BC^2 - BA^2 &=& 4\cdot AE\cdot BE \cdot \cos(\epsilon ) \quad & | \quad EC=AE \\ \hline \end{array} \)
\(\begin{array}{|rcll|} \hline R &=& R_1 + R_2 \quad & | \quad R_1 = \text{area}~\triangle AEB,\quad R_2 = \text{area}~\triangle ECB \\ &=& \dfrac{AE\cdot BE\cdot \sin(\epsilon) }{2} + \dfrac{EC\cdot BE \sin(180^\circ-\epsilon) }{2} \quad & | \quad EC=AE \\ &=& \dfrac{AE\cdot BE\cdot \sin(\epsilon) }{2} + \dfrac{AE\cdot BE \sin(\epsilon) }{2} \\ &=& AE\cdot BE\cdot \sin(\epsilon) \\\\ AE\cdot BE &=& \dfrac{R}{\sin(\epsilon)} \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline BC^2 - BA^2 &=& 4\cdot AE\cdot BE \cdot \cos(\epsilon ) \quad | \quad AE\cdot BE = \dfrac{R}{\sin(\epsilon)} \\ BC^2 - BA^2 &=& 4\cdot \dfrac{R}{\sin(\epsilon)} \cdot \cos(\epsilon ) \\ BC^2 - BA^2 &=& 4R \cdot \cot(\epsilon ) \\\\ \mathbf{\cot(\epsilon )} &=& \mathbf{\dfrac{BC^2 - BA^2}{4R}} \\ \hline \end{array} \)
