A special case of Miquel's theorem. Alternatively, let $O$ be the circumcenter of $ABC$, then notice that $AEOF$ is cyclic since $\angle AEO = \angle AFO = 90^\circ$. Similarly, $BFOD$ and $CDOE$ are also cyclic so those three circumcircles intersect at one point, which is $O$.

Lemma (Miquel Point of a Triangle). Let $D,E,F$ be points on $BC, CA$ and $AB$. Then there exists a point lying on the circumcircles of triangles $AEF, BFD$ and $CDE$. Proof. $\measuredangle ECD=\measuredangle ACB=\measuredangle EKD$ and $\measuredangle DBF =\measuredangle CBA =\measuredangle DKF$. Notice that $\measuredangle EKD + \measuredangle DKF + \measuredangle FKE = 0 \iff \measuredangle FKE = -\measuredangle EKD -\measuredangle DKF = -\measuredangle ACB - \measuredangle CBA$. But $\measuredangle BAC = -\measuredangle ACB - \measuredangle CBA$. So $\measuredangle FKE =\measuredangle BAC$. $\square$ Now the problem is utterly vanquished!

Construct circles centered at the vertices of $\triangle ABC$ with radius $0$. The pairwise radical axes of these circles are the perpendicular bisectors of the sides of $\triangle ABC$. They meet at the radical center, $O$, of the three circles, a point where $OA^2=OB^2=OC^2$. This is obviously the circumcenter. Now, with all the right angles created by the perpendicular bisectors, it is easy to check that $AEOF$, $BFOD$, and $CDOE$ are all cyclic meaning that the circumcircles of triangles $AEF, BFD$ and $CDE$ intersect all in one point, namely the circumcenter, which is what we wanted.

what was the point of constructing the radius 0 circles again