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Mathematics interesting reasoning problem
The solution ≈AEF is the exterior angle of △EBG (known).

∴∠AEF=∠B+∠G (one outer angle of a triangle is equal to the sum of two non-adjacent inner angles)

∫∠ACB is the outer corner of △FCG (known).

∴∠ACB=∠CFG+∠G (one outer angle of a triangle is equal to the sum of two non-adjacent inner angles)

∠∠CFG =∠AFE (definition of vertex angle)

∠∠AEF =∠AFE (known)

∴∠CFG =∞∠AFE =∞∠AEF (equivalent substitution)

∴∠ACB=∠AEF+∠G (equivalent substitution)

∠∠AEF =∠b+∠G (the outer angle of a triangle is equal to the sum of two non-adjacent inner angles)

∴∠ACB=∠B+∠G+∠G (equivalent substitution)

2∠G=∠ACB-∠B

That is, ∠G= 1/2(∠ACB-∠B).

It is not easy to type so many words. Give me the best answer.