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# Geometry

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• #62045
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In isosceles triangle ABC perpendicular, HE is dropped from the midpoint H of BC to side AC. Prove that if 0 is the midpoint of HE, then lines AO and BE are perpendicular

#62065
Saumik Karfa
Moderator

will get back to you soon

#62475
KOUSHIK SOM
Participant

Let A BE THE POINT (0,0) and AH the positive X axis .If C is $(x_0,y_0)$ then B has to be $(x_0,-y_0)$ and H,$(x_0,0)$

Let E=$(e_1,e_2)$,E must satisfy the equation $(\frac{e_1}{x_0}=\frac{e_2}{y_0})$ and $(\frac{e_2 -0}{e_1 -x_0})(\frac{y_0}{x_0})$=-1

i.e$(e_1y_0 = e_2 x_0)$ and $e_2y_0 = {x_2}^2 x_0 e_1$

From these we get $(e_1 x_0)y_0 = e_2{x_0}^2$

$\Rightarrow y_0({x_0}^2- e_2 y_0)=e_2 {x_0}^2$

$\Rightarrow e_2 = \frac{y_0{x_0}^2}{{x_0}^2+{y_0}^2}$

$\Rightarrow e_1 = \frac{e_2 x_0}{y_0}=\frac{{x_0}^3}{{x_0}^2+{y_0}^2}$

Hence, O=$(\frac{1}{2} (\frac{{x_0}^3}{{x_0}^2+{y_0}^2} +x_0)$,$\frac{1}{2} \frac{y_0{x_0}^2}{{x_0}^2+{y_0}^2})$

To prove that AO perpendicular BE we must check $(\frac {y_0 {x_0}^2}{2{x_0}^2+x_0{y_0}^2})(\frac {\frac {y_0 {x_0}^2}{{x_0}^2+{y_0}^2} +y_0}{\frac {{x_0}^3}{{x_0}^2+{y_0}^2 }-x_0})=-1$

L.H.S,

$(\frac{y_0 x_0}{2{x_0}^2+{y_0}^2})(\frac{ y_0{x_0}^2+y_0 {x_0}^2+{y_0}^3}{{x_0}^3-{x_0}^3-x_0{y_0}^2})$

=$(\frac {y_0 x_0}{2{x_0}^2 +{y_0}^2})(\frac{2{x_0}^2 +{y_0}^2}{-x_0 y_0})=-1$

• This reply was modified 1 month, 3 weeks ago by KOUSHIK SOM.
• This reply was modified 1 month, 3 weeks ago by KOUSHIK SOM.
• This reply was modified 1 month, 3 weeks ago by KOUSHIK SOM.
• This reply was modified 1 month, 3 weeks ago by KOUSHIK SOM.
• This reply was modified 1 month, 3 weeks ago by KOUSHIK SOM.
• This reply was modified 1 month, 3 weeks ago by KOUSHIK SOM.
• This reply was modified 1 month, 2 weeks ago by Saumik Karfa.
#63342
Participant

Except using coordinate geometry is there any other way ?

#63489
KOUSHIK SOM
Participant

Join A,D .Note that $AD \perp BC$.Also let $AF \cap BE$ be ${G}$ and $H$ be the mid point of the segment  $EC$.It suffices to prove that A,B,D,G are concyclic(because that implies $\angle ADB=\angle AGB$.This is equivalent to proving that $\angle FAD=\angle EBD$.

Join D,H .Note that $\triangle ADE |||\triangle CDE$ and the sides $DE$ and $EC$ correspond. As AF and DH are medians DE and EC respectively,they correspond too.Hence $\triangle DAF ||| \triangle DCH$

Hence $\angle FAD =\angle HDC$ As $D,H$ are mid points of $BC,CE$ respectively.we have $DH ||BE$.Hence $\angle HDC =\angle EBD$.Hence $\angle FAD=\angle EBD$

• This reply was modified 1 month, 2 weeks ago by KOUSHIK SOM.
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