Let Q be a point on side BC of triangle ABC.

Construct points P and R on sides AB and AC
respectively such that PQR is a right angle
and PR is parallel to BC.

Segment PR is a diameter of a circle with points P, Q, and R.  The center of that circle then bisects PR, and since PR is parallel to BE, the center lies on the median from vertex A to side BC.

If I knew the radius, then I cound draw the median and draw a circle centered at Q.  The intersection inside the triangle would be the center of the circle containing P, Q, and R.  That circle would intersect AC and BC at P and R.