Problem: Let ABC be a triangle with D as the midpoint of side AB, E as the midpoint of side BC, and F as the midpoint of side AC. Let k₁ be the circle passing through points A, D and F; let k₂ be the circle passing through points B, E and D; Let k₃ be the circle passing through points C, F and E. Prove that k₁, k₂ and k₃ intersect in a point.

Solution: Let the midpoints of AB, BC, and AC be D, E and F respectively. Then △ADF, △FEC, △DBE and △DFE are all congruent, and each is similar (with similarity ratio 1∕2) to the original △ABC. The circle k₁ that passes through points A, D, and F is the circumscribing circle for △ADF and similarly for the circles k₂ and k₃ through D, B, E and E, F, C respectively. Since the triangles are all congruent, the circumscribing circles are all have the same diameter.

The circumcenter of △DBE is at the intersection of the perpendicular bisectors of DB, BE, ED. Let I be the circumcenter of △DBE, and let H be the opposite end of the diameter of k₂ through B.

Now consider the diameter of k₂ through E. Let the opposite end of the diameter on k₂ be H₁. By congruence, △DBE, its circumcicle and the diameter through E can be rigidly translated to △FEC, and the diameter will pass through H. Similarly for △ADE and the diameter through A also passing through H. Hence the circles have the point H in common. Hence the three circumcircles have the point H in common.

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Commentary: This is a special case of Miquel’s Theorem and the Pivot Theorem.