About isosceles triangles

Question

Let $ABC$ be an acute-angled triangle in which $\hat{ABC}$ is the largest angle. Let $O$ be its circumcenter. The perpendicular bisectors of $BC$ and $AB$ meet $AC$ at $X$ and $Y$ respectively. The internal bisectors of $\hat{AXB}$ and $\hat{BYC}$ meet $AB$ and $BC$ at $D$ and $E$ respectively. How do I prove that $BO$ is perpendicular to $AC$ if $DE$ is parallel to $AC$.

Answer 1

“$YK$ is the perpendicular bisector of $AB$” implies $\alpha = \alpha_1$. Together with $\beta = \beta_1$, we have $AB // YE$. Similarly, $BC // DX$. as hinted by @HoseynHeydari .

Together with the given parallels, we have $AQEY$ and $XDEC$ as parallelograms.

In particular, $\beta = \gamma$ means $B, D,Y, E$ are con-cyclic. $B, D, X, E$ are con-cyclic for the same reason.

This means $DXYE$ is a cyclic quadrilateral and therefore $\omega = \beta_1$.

That is, all the red marked angles are equal to all the green marked angles.

∴ $\triangle BAC$ is isosceles with $BA = BC$. If $O$ is the circum-center, then $BO$ is the third perpendicular bisector besides $XH$ and $YK$.

Result follows.