Prove:

Assume that △ABC is not an equilateral triangle under the condition of the topic.

At this time, we might as well set the side length to satisfy the relationship a>= c & gt=b, while a>b

∴∠a>； =∠C & gt； = < b and < ∠A>；; ∠B

∴0 <∠B& lt； ∠A & lt； 180

∴3∠a>； ∠ A+∠ B+∠ C = 180, namely: ∠ A > 60

3∠B& lt； ∠ A+∠ B+∠ C = 180, namely: ∠ B.

∫60+∠BEF & gt； ∠B+∠BEF=∠AFE=∠DFE+∠AFD=60 +∠AFD

∴∠bef>； ∠AFD

Situation 1:

If af = be

Then according to the big side and big angle of the triangle:

∠BFE & lt； =∠B

∴2∠bfe<； =∠BFE+∠B& lt； 180

∴∠bfe<； 90

You can draw the same conclusion: ∠ FDA < 90

Using sine theorem in △BEF and △AFD;

sin∠BFE/sin∠B = BE/EF = AF/FD = sin∠FDA/sin∠A

∴sin∠bfe/sin∠fda=sin∠b/sin∠a=b/a<； 1

∴sin∠bfe<； Sin∠FDA and ∠BFE and ∠FDA are all acute angles.

∴∠bfe<； United States Food and Drug Administration

∴∠BFA

=∠BFE+∠DFE+∠DFA

=∠BFE+60+∠DFA & lt； ∠FDA+∠A+∠DFA= 180

∠BFA is not 180, B, F and A are not in a straight line, which is contradictory!

Situation 2

If AF = BE & gt=DF=EF.

∵ 180 & gt； ∠BEF & gt； ∠AFD & gt； 0

∴cos∠bef<； cos∠AFD

Substitute cosine theorem into the above formula:

(be^2+ef^2-bf^2)/(2be*ef)<； (AF^2+FD^2-AD^2)/(2AF*FD)

∴be^2+ef^2-bf^2<； AF^2+FD^2-AD^2

∴bf^2>； AD^2

∴bf>； advertisement

BFE

=(BF^2+EF^2-BE^2)/(2BF*EF)

=[BF-(BE^2-EF^2)/BF]/2EF

& gt[AD-(BE^2-EF^2)/AD]/2EF

=[AD-(AF^2-FD^2)/AD]/2FD

=(AD^+FD^2-AF^2)/(2AD*FD)

=cos∠FDA

∴∠bfe<； United States Food and Drug Administration

At this time, the same situation appeared.

∠BFA

=∠BFE+∠DFE+∠DFA

=∠BFE+60+∠DFA & lt； ∠FDA+∠A+∠DFA= 180

∠BFA is not 180, B, F and A are not in a straight line, which is contradictory!

∴ The assumption is not true, that is, △ABC is an equilateral triangle.