∴∠EGC=∠ADC=90
∴EG∥AD
∴∠E=∠DAC
Also ∫ known ∠E =∠ bad
∴∠BAD=∠DAC
Ad Split ∠BAC
(2) Judgment: GF⊥AB
Reason: ∵AC⊥BC DE⊥AC
∴∠3+∠DCE=90 ∠ 1+∠DCE=90
∴∠ 1=∠3
∵∠ 1=∠2
∴∠3=∠2
∴CD∥GF
∵CD⊥AB
∴GF⊥AB