+302 
In the figure above, DF is the diameter of the semicircle. Rectangle ABCD and BEFG are congruent, and EF is on the diameter DF. Find the measure of ∠ABG.
+9673 Note that since the rectangles are congruent, we have DB = BF. That would imply \(\angle BDF = \angle BFD\). Also, BE is perpendicular to EF, so \(\angle BED = \angle BEF = 90^\circ\). Together with the common side BE we have the pair of congruent triangles \(\triangle DEB \cong \triangle FEB\). Consequently we have the fact that E is the centre of the semi-circle. So we now have enough information to figure out that \(\triangle DAE\) is equilateral. The rest is done through some angle chasing, which is not hard.
