Known: as shown in the figure, in square ABCD, e is the midpoint of BC, point F is on CD, angle FAE is equal to angle BAE, AF = BC + EC

Known: as shown in the figure, in square ABCD, e is the midpoint of BC, point F is on CD, angle FAE is equal to angle BAE, AF = BC + EC

Do eg ⊥ AF to g, connect EF
∵∠ABE=∠AGE=90°,∠FAE=∠BAE
AE=AE
∴△ABE≌△AGE(AAS)
∴AG=AB=BC
BE=EG
∵ e is the midpoint of BC, then be = CE = eg
EF=EF
∴RT△EFG≌RT△EFC(HL)
∴FG=CF
∴AF=AG+FG=BC+CF

Known: as shown in the figure, in square ABCD, e is the midpoint of BC, point F is on CD, ∠ FAE = ∠ BAE

No graph, are you sure it's in a square? Without graph, I suppose your side length is 2, BC + EC = 3
What about AF? It floats between 2 and 2 root 2. Unfortunately, there is no map

As shown in the figure, EF is the vertical bisector of the diagonal BD of the parallelogram ABCD. EF intersects the edges AD and BC at e and f respectively

The ∵ quadrilateral ABCD is a parallelogram ∵ ad ∥ BC (the opposite sides of parallelogram are parallel) ∵ AEF = ∠ CFE (two lines are parallel and the internal stagger angles are equal) ∵ EF vertical bisector diagonal AC ∵ Ao = Co, EF ⊥ AC (the definition of the vertical bisector of a line segment) ∵ Ao = Co, EF ⊥ AC, ∠ AEF = ∠ CFE ≌ △ AOE ≌ △ COF (the congruence of two triangles corresponding to two equal angles and the opposite sides of one angle) ≌ of = OE ∵ EF ⊥ AC, Of = OE, Ao = Co  quadrilateral aecf is rhombic (the diagonals of quadrilateral are perpendicular to each other, bisection is rhombic). \ \ x0d to judge a quadrilateral is rhombic, we can prove that it is parallelogram from side, angle and diagonal, and then prove that it is rhombic, or prove that the four sides are equal to get rhombic directly, For example, it can be proved that two diagonals are perpendicular to each other to prove that they are rhombus, or that a group of opposite sides are parallel and equal, and a group of adjacent sides are equal, etc. when diagonals are perpendicular to determine that a quadrilateral is rhombus, we must first prove that the quadrilateral is parallelogram, that is, only when the conditions are sufficient can we draw a conclusion, As long as you master the method, it will be easy to solve similar problems

(2008 · Guang'an) as shown in the figure, in ladder ABCD, ad ‖ BC and E are the midpoint of CD, connecting AE and extending the extension line between AE and BC at point F
(1) (2) if ad = 2, ab = 8, when BC is, point B is on the vertical bisector of line AF, why?

(1) It is proved that: ∵ ad ∥ BC, ∵ f = ∠ DAE. (1 point) and ∵ FEC = ∠ AED, ∵ ECF = ∠ ade, ∵ e is the midpoint of CD, ∵ CE = de. in △ FEC and △ AED,