Let $M=XE\cap BC , N=XF\cap BC$ $\angle{XCF}=\angle{CAO}+\angle{CXO}$ $=\angle{CAO}+\angle{CBO}$ $=\angle{C}$ $=\angle{NCF}$ So $F$ is the midpoint of $XN$ . similary , $E$ is the midpoint of $XM$ Therefore $A$ is the circumcenter of $\Delta XMN$ so $D$ is the midpoint of $MN$ and by taking homothity we are done. Remark: I think it could be done by complex bash but I couldn't compute $x$ @below Nice idea

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For complex bash you could just say that x=ta and t is real and then compute all other points is simple and assume x,d and the midpoint of EF are collinear and this way you find t. Now only need to show that XOBC lie on a common circle which is rather simple

Note that $\triangle AEF\sim\triangle ABC$ and by law of sine we have $$\frac{CF}{CA}=\frac{CF}{CX}\cdot\frac{CX}{OC}\cdot\frac{OC}{CA}=\cos C\cdot\frac{\sin 2B}{\cos A}\cdot\frac{1}{2\sin B}=\frac{\cos C \cos A}{\cos B}.$$ Note that $\displaystyle \frac{HD}{AH}= \frac{\cos C \cos A}{\cos B}$ by direct computation, where $H$ is the orthocenter of $\triangle ABC$. This implies $D$ is the orthocenter of $\triangle AEF$ and by using a lemma on the orthocenter and the circumcircle of $\triangle AEF$, we are done.