∫BD divided equally ∠ABC FG⊥BC

∴∠BFG=90 -？ ∠ABC=67.5

∠∠ABC = 45∠ACB = 65

∴∠BAC=70

∫∠BFH is the outer corner of △ABF.

∴∠BFH=？ ∠ABC+？ ∠BAC=57.5

∴∠hfg=∠bfg-∠bfh=67.5-57.5 = 10

(2)

According to the idea in (1)

∠BFG=90 -？ ∠ABC

∠BFH=？ ∠ABC+？ ∠BAC

=? ∠ABC+？ ( 180 -∠ABC-∠ACB)

=90 -? ∠ACB

∴∠HFG=∠BFG-∠BFH

=? (∠ACB-∠ABC)

(3)BFH =∠CFG

Reason:

∫∠BFH is the outer corner of △ABF.

∴∠BFH=？ (∠ABC+∠BAC)

In the right triangle FGC,

∠CFG=90 -？ ∠ACB

=90 -? ( 180 -∠ABC-∠BAC)

=? (∠ABC+∠BAC)

Therefore, ∠BFH=∠CFG.

The method may not be the simplest, and the answer is for reference only.

It's done.