Dear ╭(╯3╰)╮, please choose the satisfactory answer ~ ~.

Dear ╭(╯3╰)╮, please choose the satisfactory answer ~ ~.

In △DEF, DE=DF, and a straight line passing through point A on EF intersects with the extension lines of DE and DF at point B and point C respectively, and be = CF 。

Proof: ab = AC.

Testing Center: congruent triangles's Judgment and Nature.

Special topic: proving the problem.

Analysis: Let B be BG∨CD, EF and G intersect. According to the proportion theorem of parallel lines, there is BG: DF = BE: DE, and known DE=DF, so BG=BE, and BE=CF, then the equivalent substitution is BG=CF, which is easy to prove △ ACF △ ABG, and according to the properties of congruent triangles, there is AB =

Answer: proof: pass b as BG∨CD and EF to g,

∫BG∨CD，

∴BG:DF=BE:DE,∠AGB=∠AFC，

And DE = DF,

∴BG=BE，

And ∵BE=CF,

∴BG=CF，

Similarly, ≈GAB =∠FAC,

∴△ACF≌△ABG，

∴AB=AC.

Comments: This question examines congruent triangles's judgment and nature; In solving the problem, we use the theorem of proportion of parallel lines, congruent triangles's judgment and properties, and equivalent substitution. Making auxiliary lines correctly is the key to solve this problem.

Dear ╭(╯3╰)╮, please choose the satisfactory answer ~ ~.

Dear ╭(╯3╰)╮, please choose the satisfactory answer ~ ~.

Dear ╭(╯3╰)╮, please choose the satisfactory answer ~ ~.