As can be seen from the figure 1, O is the midpoint of BC and is connected with AO.

Then AO=OC, ∠ Bao = 45.

∠∠FOC = 180-∠EOF-∠EOB

∠ EOF = 90 ∠ FOC = 90-∠ EOB。

∠ AOE = 90-∠ EOB。

∴∠AOE=∠FOC

∠∠boa =∠foc = 45。

AO=OC

∴△AOE≌△COF

∴AE=FC。

The second question: first prove the similarity, and then calculate it in proportion.

Connecting AO

∠FOC=90 -∠AOF=90 -(∠EOF-∠AOE)

=45 +∠AOE

∠∠ Bao = 45, ∠ Bao +∠AOE=∠BEO.

∴∠FOC=∠BEO

∠∠B =∠C

∴△BEO∽△COF

BE:CO=BO:CF

BE=x，CF=y

Y=2/x, (0 < x