解決方案:
∠∠ACF =∠DCF
∠ECB=∠ACE
∠∠ACF+∠DCF+∠ECB+∠ACE = 180度。
∴∠ECF=∠ACE+∠ACF=90學位
∠∠ACD =∠A+∠ABC
還有≈ABF+∠A =∠ACF+∠f
∠ACF=∠DCF=1/2 ∠ACD
∠ABF=∠FBD=1/2 ∠ABC
因此∠F=∠ABF+∠A-∠ACF。
= 1/2∠ABC+∠A-1/2∠ACD
= 1/2∠ABC+∠A-1/2(∠A+∠ABC)
=1/2∠A
=30度