sqing wrote: Let $ABC$ be an isosceles triangle with $AB=AC.$ Let $ \Gamma $ be its circumcircle and let $O$ be the centre of $ \Gamma $ . let $CO$ meet $ \Gamma$ in $D .$ Draw a line parallel to $AC$ thrugh $D.$ Let it intersect $AB$ at $E.$ Suppose $AE : EB=2:1$ .Prove that $ABC$ is an equilateral triangle. Let $AH$ be the altitude of the triangle. Because of $DE||AC$ we have $E\widehat DA = D\widehat AC = {90^0}$ and $D\widehat EA = \widehat A$. But, $A\widehat OD = O\widehat AC + O\widehat CA = \widehat A = D\widehat EA$, thus $AOED$ is cyclic and $EO||BC$, so $\frac{{AO}}{{OH}} = \frac{{AE}}{{EB}} = \frac{2}{1}$, that means $O$ is centroid. In the isosceles triangle $ABC$, circumcenter $O$ is also centroid, hence the triangle is equilateral.

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george_54 wrote: sqing wrote: Let $ABC$ be an isosceles triangle with $AB=AC.$ Let $ \Gamma $ be its circumcircle and let $O$ be the centre of $ \Gamma $ . let $CO$ meet $ \Gamma$ in $D .$ Draw a line parallel to $AC$ thrugh $D.$ Let it intersect $AB$ at $E.$ Suppose $AE : EB=2:1$ .Prove that $ABC$ is an equilateral triangle. Let $AH$ be the altitude of the triangle. Because of $DE||AC$ we have $E\widehat DA = D\widehat AC = {90^0}$ and $D\widehat EA = \widehat A$. But, $A\widehat OD = O\widehat AC + O\widehat CA = \widehat A = D\widehat EA$, thus $AOED$ is cyclic and $EO||BC$, so $\frac{{AO}}{{OH}} = \frac{{AE}}{{EB}} = \frac{2}{1}$, that means $O$ is centroid. In the isosceles triangle $ABC$, circumcenter $O$ is also centroid, hence the triangle is equilateral. Wonderful. Thanks.

Easy to see, $\Delta AOB \cong \Delta AOC$ $\implies$ $\angle ABO=\angle ACO$, Since, $\angle DAC=90^{\circ}$ $\implies$ $\angle EDA=90^{\circ}$, hence, if $A'$ is the $A-$ antipode in $\odot (ABC)$, then, $A' \in ED$, $\angle DA'A=\angle DCA=\angle EBO$ $\implies$ $EBA'O$ is cyclic, $\implies$ $EO||BC$ hence, if $G$ is the centroid of $\Delta ABC$, then, $O \equiv G$ $\implies$ $\Delta ABC$ is equilateral

At last solved a RMO geometery problem...I am a geo noob

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