Question

并列三个边长相同的正方形ABCD,正方形CDEF,正方形EFGH.求证：∠ACB+∠AFB+∠AGB=90°

Answer 1

设AB＝1

.则CF∶AC∶FA＝1∶√2∶√5＝√2∶2∶√10＝AC∶CG∶AG

∴⊿FCA∽⊿ACG

∠AGB＝∠CAF

∠ACB+∠AFB+∠AGB=45º+∠AFB+∠CAF＝45º+∠ACB﹙外角﹚＝45º+45º＝90º