The figure shows a square ABCD. Arc BD with center at A and AC intersect at F. The extension of DF meets BC at G. Prove that BG = 2.EF. Post a comment or solution.
 

See Also:

Proposed Problem 395.
Square, 15 Degree, Equilateral triangle.

Proposed Problem 393.
Triangle, Orthocenter, Circumcircles, Congruence, Collinear.

Proposed Problem 392.
Triangle, Parallel lines, Concurrent lines.