解:(1)∵∠ABC+∠DCB=360°﹣(α+β), ∴∠ABC+(180°﹣∠DCE)=360°﹣(α+β)=2∠FBC+(180°﹣2∠DCF)=180°﹣2(∠DCF﹣∠FBC) =180°﹣2∠F, ∴360°﹣(α+β)=180°﹣2∠F, 2∠F=α+β﹣180°, ∴∠F= (α+β)﹣90°; (2)∵∠ABC+∠DCB=360°﹣(α+β), ∴∠ABC+(180°﹣∠DCE)=360°﹣(α+β)=2∠GBC+(180°﹣2∠HCE)=180°+2(∠GBC﹣∠HCE) =180°+2∠F, ∴360°﹣(α+β)=180°+2∠F, ∠F=90°﹣(α+β); (3)α+β=180°时,不存在∠F. |