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.