Equal segments AF and CG are plotted on the diagonal AC of rectangle ABCD. Prove that BGDF is a parallelogram.

1. Let’s draw a drawing.

2. Let us prove that DF = GB.

Consider triangles ADF and CBG. These triangles are equal in two sides and the angle between them:

AD = BC as opposite sides of the rectangle;

AF = CG – by the condition of the problem;

∠DAF = ∠BCG, as crosswise lying with parallel AD and BC and secant AC.

Hence, DF = GB.

3. Let us prove that BF = DG.

Consider triangles ABF and CDG. These triangles are equal in two sides and the angle between them:

AB = DC, as opposite sides of the rectangle;

AF = CG – by the condition of the problem;

∠BAF = ∠DCG, as internal criss-crossing with parallel AB and DC and secant AC.

Hence, BF = DG.

4. Let us prove that BGDF is a parallelogram.

A quadrilateral is a parallelogram if its sides are equal in pairs. In our case, this condition is met:

DF = GB from triangles ADF and CBG;

BF = DG from triangles ABF and CDG.

Hence BGDF is a parallelogram.