Acute triangle △ABC satis es AB<AC. Let the circumcircle of this triangle be O, and the midpoint of BC,CA,AB be D,E,F. Let P be the intersection of the circle with AB as its diameter and line DF, which is in the same side of C with respect to AB. Let Q be the intersection of the circle with AC as its diameter and the line DE, which is in the same side of B with respect to AC. Let PQ∩BC=R, and let the line passing through R and perpendicular to BC meet AO at X. Prove that AX=XR. geometrycircumcircleperpendicularequal segments