In the parallelogram ABCD (Xining City, 2008), the bisector ce of angle BCD intersects the edge ad at E (Xining City, 2008) 23. As shown in Figure 10, it is known that in the parallelogram ABCD, the bisector ce of angle BCD intersects ad at e, the bisector BG of angle ABC intersects CE at F, and ad at g. verification: AE = DG

In the parallelogram ABCD (Xining City, 2008), the bisector ce of angle BCD intersects the edge ad at E (Xining City, 2008) 23. As shown in Figure 10, it is known that in the parallelogram ABCD, the bisector ce of angle BCD intersects ad at e, the bisector BG of angle ABC intersects CE at F, and ad at g. verification: AE = DG

prove:
∵ in parallelogram ABCD, ad ‖ BC, ab = CD
∴∠AGB=∠CBG,∠CED=∠BCE
∵ BG bisection ∠ ABC, CE bisection ∠ BCD
∴∠ABG=∠CBG,∠DCE=∠BCE
∴∠AGB=∠ABG,∠CDE=∠DCE
∴AB=AG,CD=DE
∴AG=DE
∴AG-EG=DE-EG
That is AE = DG