Prove that the line joining the centroid and the incenter of a non-isosceles triangle is perpendicular to the base if and only if the sum of the other two sides is thrice the base.
CatalinBordea wrote:
Let be a non-isosceles triangle $ ABC $ having $ G $ as its center of mass, and $ I $ as its incircle. Prove that $ GI\perp BC $ if and only if $ AB+AC=3BC. $
$D$ be touch-point of the incircle with $BC$.Then $GD\perp BC\implies GB^2-GC^2=DB^2-DC^2$.Now $$\dfrac{2c^2+2a^2-b^2}{9}-\dfrac{2b^2+2a^2-c^2}{9}=GB^2-GC^2=DB^2-DC^2=(s-b)^2-(s-c)^2=a(c-b)\Rightarrow b+c=3a$$.