Geometry e is a point of ad on the edge of square ABCD, BF bisects ∠ EBC intersects DC with F, and proves that EB = AE + CF

Geometry e is a point of ad on the edge of square ABCD, BF bisects ∠ EBC intersects DC with F, and proves that EB = AE + CF

Since Ag = FC, Ba = BC, gab = FCB = 90, AGB and BFC are congruent
So GBA = FBC, BGA = BFC
Because AB / / CD, ABF = BFC, BGA = ABF,
Because BF bisects ∠ EBC, EBF = FBC, and GBA = FBC, EBF = GBA, so EBF + Abe = GBA + Abe, GBE = ABF
Combining with BGA = ABF, we get BGA = GBE
So EGb is isosceles triangle, be = Ge
Ge = AE + GA = AE + CF
So be = AE + CF