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I am trying to prove the following:

Let $T$ be the Gergonne point (the intersection of the lines that connect the points of tangency of the incircle with the vertices of the triangle) of $\triangle ABC$. Show that if $T$ coincides with the incenter or the circumcenter or the orthocenter or the centroid of $\triangle ABC$, then the triangle must be equilateral.

I would appreciate any help. Thank you!

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Case 1

$T=H$ coincides with orthocenter of $ABC.$ Let $T_a,$ $T_b,$ $T_c$ be the points where incircle touches $BC,$ $AC$ and $AB$ respectively. Since $TT_a$ is perpendicular to $BC,$ $TT_b$ is perpendicular to $AC,$ $TT_c$ is perpendicular to $AB$ it follows that $T$ is also an incenter of $ABC.$ Hence, all bisectors are simultaneously altitudes of the triangle which implies that $ABC$ is equilateral.

Case 2

$T=M$ coincides with a centroid of $ABC.$ Then, $BT_a=CT_a$ or using well-known formulas $BT_a=\frac{a+c-b}{2}=CT_a=\frac{a+b-c}{2}.$ Last equality implies that $b=c.$ In exactly the same way one can show that $a=c.$ Hence, $ABC$ is again equilateral.

Case 3

$T=O$ coincides with the circumcenter of $ABC.$ IMHO, the most natural way is to "compute" a bit. More precisely, $\angle T_cAB=90-\angle ACB=90-\gamma,$ $\angle T_aAC=90-\angle ABC=90-\beta.$ Note that the ratio of the areas of $\triangle AT_aB$ and $AT_aC$ is equal to $$\frac{BT_a}{CT_a}=\frac{p-b}{p-c}=\frac{AB\cos\gamma}{AC\cos\beta}=\frac{c\cos\gamma}{b\cos\beta}=\frac{c^2(b^2+a^2-c^2)}{b^2(a^2+c^2-b^2)}.$$

After that one can get the same expressions for the other sides and play with equalities a little bit to get that $a=b=c.$

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