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I understand by the Euler line that the centroid, circumcenter, and orthocenter are collinear, but I don't know how to fit in the fact about the incenter and the isosceles triangle

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2 Answers 2

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  • in $\bigtriangleup$ABC AB=AC
  • WE take AD$\perp$BC.clearly BD=DC & $\angle$BAD=$\angle$DAC
  • so clearly the circumcentre;orthocentre;incentre and centroid - all of them lie on the line AD
  • SO that is proovedenter image description here
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Given the constructions of these centers $c_i$, a congruence $\Delta\to\Delta'$ of two triangles will transport $c_i$ to $c_i'\ $. As an isosceles triangle is congruent to its mirror image we have $c_i=c_i'$ for each of these centers. therefore they all lie on the symmetry axis.

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