Triangle ABC is an isosceles right-angled triangle (angle C = 90). The midpoints of the sides AB, BC

Triangle ABC is an isosceles right-angled triangle (angle C = 90). The midpoints of the sides AB, BC, CA are designated by the points D, E, F, respectively. The segments DC, DE, DF are drawn. Prove that point D will be equidistant from the vertices of the given triangle.

AF = FC = CE = EB, AD = DB, since triangle ABC is isosceles, and points F, E, D are the midpoints of the sides. СD – median, bisector and height of triangle ABC. Consequently, the angles DCB, CBD, DAC, ACD are equal to 45 ° (angles B and A are equal, since they are located at the base of an isosceles triangle). From all this it follows that the triangles ACD and DCB are isosceles, CD = DB = DA.