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  • 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  

 Jul 10, 2019
edited by OldTimer  Jul 10, 2019
 #1
avatar+103689 
+2

Can we assume that  4R is meant to be in brackets?

 Jul 10, 2019
 #2
avatar+103689 
+2

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

 

\(R=0.5zwsin\theta+0.5zwsin(180-\theta)\\ R=0.5zwsin\theta+0.5zwsin(\theta)\\ R=zwsin\theta\\ 4R=4zwsin\theta\\ \)

 

\(y^2=w^2+z^2-2wzcos(180-\theta)\\ y^2=w^2+z^2+2wzcos(\theta)\\ x^2=w^2+z^2-2wzcos(\theta)\\ subtracting\\ y^2-x^2=4wzcos(\theta)\\ BC^2-BA^2=4wzcos(\theta)\\ \)

\(\therefore\\ RHS =\frac{BC^2-BA^2}{4R}\\ RHS =\frac{4wzcos(\theta)}{4wzsin(\theta)}\\ RHS =cot(\theta)\\ RHS=LHS \qquad QED\)

.
 Jul 10, 2019
 #4
avatar+142 
+2

Thanks..its a joy to see these solutions! Regards

OldTimer  Jul 11, 2019
 #6
avatar+103689 
+2

You are always welcome.  laugh

Melody  Jul 11, 2019
 #3
avatar+23071 
+3

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:

\(\cot(AEB) = \dfrac{ BC^2 - BA^2 } { 4R } \)

 

\(\text{Let $\angle AEB= \epsilon$} \\ \text{Let $AE=EC$} \)

 

\(\begin{array}{|rcll|} \hline BC^2 &=& AE^2 + BE^2 - 2\cdot AE\cdot BE \cdot \cos(180^\circ-\epsilon ) \\ &=& AE^2 + BE^2 + 2\cdot AE\cdot BE \cdot \cos(\epsilon ) \\\\ BA^2 &=& EC^2 + BE^2 - 2\cdot EC\cdot BE \cdot \cos(\epsilon ) \\\\ BC^2 - BA^2 &=& 4\cdot AE\cdot BE \cdot \cos(\epsilon ) \quad & | \quad EC=AE \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline R &=& R_1 + R_2 \quad & | \quad R_1 = \text{area}~\triangle AEB,\quad R_2 = \text{area}~\triangle ECB \\ &=& \dfrac{AE\cdot BE\cdot \sin(\epsilon) }{2} + \dfrac{EC\cdot BE \sin(180^\circ-\epsilon) }{2} \quad & | \quad EC=AE \\ &=& \dfrac{AE\cdot BE\cdot \sin(\epsilon) }{2} + \dfrac{AE\cdot BE \sin(\epsilon) }{2} \\ &=& AE\cdot BE\cdot \sin(\epsilon) \\\\ AE\cdot BE &=& \dfrac{R}{\sin(\epsilon)} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline BC^2 - BA^2 &=& 4\cdot AE\cdot BE \cdot \cos(\epsilon ) \quad | \quad AE\cdot BE = \dfrac{R}{\sin(\epsilon)} \\ BC^2 - BA^2 &=& 4\cdot \dfrac{R}{\sin(\epsilon)} \cdot \cos(\epsilon ) \\ BC^2 - BA^2 &=& 4R \cdot \cot(\epsilon ) \\\\ \mathbf{\cot(\epsilon )} &=& \mathbf{\dfrac{BC^2 - BA^2}{4R}} \\ \hline \end{array} \)

 

laugh

 Jul 10, 2019
 #5
avatar+142 
+3

Thanks...works of art!

OldTimer  Jul 11, 2019
 #7
avatar+103122 
+3

THX to Melody and heureka.....!!!!

 

 

cool cool cool

CPhill  Jul 11, 2019
 #8
avatar+23071 
+3

Thank you, CPhill !

 

laugh

heureka  Jul 11, 2019
 #9
avatar+103689 
+1

Thanks Chris :)

Melody  Jul 12, 2019

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