因為 ∠AFO 為三角形BFG的外角 ,
所以上式=180°—1/2∠A—(∠FBC ∠C)
= 180°— [ 1/2∠A ( 1/2 ∠B ∠OCB ∠FCO) ]
= 180°— [ 1/2∠A ( 1/2 ∠B ∠OCB 1/2 ∠C )]
= 180°— [ 1/2∠A 1/2 ∠B 1/2 ∠C ∠OCB]
=180°— [ 1/2(∠A ∠B ∠C) ∠OCB]
=180°— [ 90° ∠OCB]
=90°—∠OCB
=∠GOC
∠AOF與∠BOD為對頂角
所以∠BOD=∠GOC