Geo2.39

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Prove that the perpendicular bisectors of the sides of a triangle are concurrent.

Let A(x_{1},y_{1}),B(x_{2},y_{2}),C(x_{3},y_{3})\, be the vertices of the triangle ABC.

If D,E,F are the midpoints of the sides BC,CA,AB\, respectively,then

D=\left({\frac  {x_{2}+x_{3}}{2}},{\frac  {y_{2}+y_{3}}{2}}\right),E=\left({\frac  {x_{3}+x_{1}}{2}},{\frac  {y_{3}+y_{1}}{2}}\right),F=\left({\frac  {x_{1}+x_{2}}{2}},{\frac  {y_{1}+y_{2}}{2}}\right)\,

Slope of BC is y_{2}-{\frac  {y_{3}}{x_{2}-x_{3}}}\,

Therefore,slope of the perpendicular bisector to the side is -{\frac  {x_{2}-x_{3}}{y_{2}-y_{3}}}\,

Equation to the perpendicular bisector thro'D is

y-{\frac  {y_{2}+y_{3}}{2}}=-{\frac  {x_{2}-x_{3}}{y_{2}-y_{3}}}(x-{\frac  {x_{2}+x_{3}}{2}}\,

Simplifying, we have

L_{1}\equiv 2x(x_{2}-x_{3})+2y(y_{2}-y_{3})-(x_{2}^{{2}}-x_{3}^{{2}}-(y_{2}^{{2}}-y_{3}^{{2}})=0\,

Similarly the other perpendicular bisectors are

L_{2}\equiv 2x(x_{3}-x_{1})+2y(y_{3}-y_{1})-(x_{3}^{{2}}-x_{1}^{{2}}-(y_{3}^{{2}}-y_{1}^{{2}})=0\,

L_{3}\equiv 2x(x_{1}-x_{2})+2y(y_{1}-y_{2})-(x_{1}^{{2}}-x_{2}^{{2}}-(y_{1}^{{2}}-y_{2}^{{2}})=0\,

Now 1.L_{1}+1.L_{2}+1.L_{3}=0\, implies L_{1}=0,L_{2},L_{3}=0\, are concurrent.

Hence the required is proved.


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