Olympic geometry problems
Solution: If D is DG perpendicular to G, it is easy to find that DG = √ 2/2 is in the isosceles right triangle DEF; Let CD=X, then AC = √ (x 2+5 2) is equal in area, ADxCD=ACxDG, if you substitute CD=X, AC = √ (x 2+5 2), AD=5, DG=√2/2, you get X=5/7. BC=2CD= 10/7 .