It is known that: as shown in the figure, in trapezoidal ABCD, ad ‖ BC, BC = DC, CF bisects ∠ BCD, DF ‖ AB, and the extension line of BF intersects DC at point E

It is known that: as shown in the figure, in trapezoidal ABCD, ad ‖ BC, BC = DC, CF bisects ∠ BCD, DF ‖ AB, and the extension line of BF intersects DC at point E


In △ BFC and △ DFC, BC = DC, BC = DC, BCF = dcffc = FC △ BFC 8780 ≌ BCF = DCF. In △ BFC and △ DFC and △ DFC, in △ BFC and △ DFC and △ DFC, BC = DC = DC · \8780; FDB. In △ BFC and △ BFC and △ DFC and 8757; BDC. ∵ BDA = ∵ BDC It is a common edge, ≌ bad ≌ bed (ASA). ≌ ad = De



It is known that: as shown in the figure, in trapezoidal ABCD, ad ‖ BC, BC = DC, CF bisects ∠ BCD, DF ‖ AB, and the extension line of BF intersects DC at point E


It is proved that: (1) in △ BFC and △ DFC, BC = DC, BCF = dcffc = FC, BFC ≌ DFC (SAS); (2) connect BD. ≌ BFC ≌ DFC, BF = DF, FBD = FDB. ∥ DF ∥ AB, abd = FDB. ≌ abd = FBD



If the internal angle a of △ ABC satisfies cosa > SINB, then the shape of △ ABC is?


Because cosa > SINB, sin (π / 2-A) > SINB. If B is an acute angle, there is π / 2-A > b, a + B