Let \(\triangle ABC\) be an acute triangle, and let \(D,E,F\) be the midpoints of \(BC,CA,AB\) respectively. Let \(A', B', C'\) be the diametrically opposite points of \(A,B,C\) respectively on the circumcircle of \(\triangle ABC.\) It turns out that lines \(A'D, B'E, C'F\) are concurrent at a point \(X\) within \(\triangle ABC.\) Then, \(X\) is the ...... of \(\triangle ABC.\)

**Details and assumptions**

- The diametrically opposite point of a point \(P\) on a circle \(\omega\) is the unique point \(P'\) also lying on \(\omega\) apart from \(P\) such that \(PP'\) is a diameter of \(\omega.\)

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