设∠ABO=α则∠CBO=α,∠BAO=θ则∠CAO=θ,∠ACO=β则∠BCO=β,∠BOD=x, ∠DOG=z,∠COG=y,∴由外角定理,直角三角形两锐角互余得：①x=α+θ②z+y=β+θ③α+x+z=90°④y+β=90°,由①②消去θ得：⑤：x－α=y+z...