As shown in the figure, in equilateral △ ABC, CD intersects be at point F, ∠ BFC = 120?: ad = CE

As shown in the figure, in equilateral △ ABC, CD intersects be at point F, ∠ BFC = 120?: ad = CE


It is proved that: ∵ BFC = 120 °
∴∠CBE+∠DCB=60°
∵ △ ABC is an equilateral triangle
∴∠A=∠ACB=∠ACD+∠DCB=60°
AC=BC
∴∠ACD=∠CBE
∵∠A=∠BCE AC=CB ∠CBE=∠ACD
∴△ACD≌△CBE(ASA)
∴AD=CE



As shown in the figure, △ ABC is an equilateral triangle, points D and E are on AB and AC respectively, and F is the intersection of be and CD. It is known that ∠ BFC = 120 °. Verification: ad = CE


It is proved that: ∵ - BFC = 120 °, ∵ - ECF = ∵ BFC - ∵ CEB = 120 ° - ∵ ABC is equilateral triangle, ∵ - EBC = 180 ° - 60 ° - ∵ CEB = 120 ° - ∵ - ECC, ∵ ECF =