∑△EBF and △EB'f overlap.

∴△EBF≌△EB'F

∴∠ 1=∠2

It can also prove:? ∠3=∠4

∵∠ 1+∠2+∠3+∠4= 180

∴2(∠2+∠4)= 180

∴∠2+∠4=90

∵? △BEF？ ∠B=90？

∴∠2+∠5=90

∴∠4=∠5? That is, ∠BEF=∠CFG.

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(2)

∫△AFH and△ a 'fh overlap.

∴△AFH≌△A'FH

∴∠ 1=∠2

It can also prove:? ∠4=∠5

∵∠ 1+∠2+∠3= 180 ∠4+∠5+∠6= 180 ?

∴∠ 1+∠2+∠3+∠4+∠5+∠6=360 ?

∴2(∠ 1+∠4)+(∠3+∠6)=360

∵△BEF？ ∠B=90

∴∠3+∠6=90

∴2(∠ 1+∠4)+90 =360

∴∠ 1+∠4= 135

∠∠EPF+∠ 1+∠4 = 180

∴∠EPF=45