Junior high school mathematics, equilateral triangles, looking forward to the master s answer, I wou

Solution: Connect DE

Because the triangle ABC is an equilateral triangle.

So the angle BAC = angle C = 60 degrees.

ab=ac=bc

Because the points are the midpoints of ab and ac, respectively.

So DE is the median line of the triangle ABC.

So de=1 2bc

ad=1/2ab

ae=1/2ac

de parallel bc

So the horn def = the horn efc

So ad=ae=de

So the triangle ade is an equilateral triangle.

So the angle ADE = angle ADQ + angle EDG = 60 degrees.

Because the triangle DFG is an equilateral triangle.

So df=dg

Angle FDG = Angle EDF + Angle EDG = 60 degrees.

So angular adg = angular edf

Because ad=de (proven).

dg = df (verified).

So triangle dag and triangle def congruence (sas) so angle dag = angle def

Because angular dag = angular bac + angular cag

Angular cag = 30 degrees.

So the angle dag = 90 degrees.

So the angle EFC = 90 degrees.

So the angle cef = 90-60 = 30 degrees.

So ce=1 2cf

ce^2=ef^2+cf^2

Because cf=1

So ce=2

So ef = root number 3

The connection de d,e is the midpoint of the equilateral triangle abc ab,ac so de=half of bc and bc=ab so de=half of ab ad=half of ab so ad=half of ab so ad=de . DE is the median line of the triangle ABC So ed parallel BC so AD=60° is equal to ADG+GDE+GDE is also equal to FDE+ GDE So FDE= ADG AD=DE=DG So the triangle ADG is all equal to the triangle EDF So def= DAG=90° And because Ed parallel BC so EFC=90° C=60° According to the special angle theorem, EF=cf 3 times the root number, so EF=root number 3 Pure hand.

Solution: Connect DE

ABC is an equilateral triangle.

ab=ac=bc

b=∠c=∠bac=60°

Points d and e are the midpoints of ab and ac, respectively.

ad=1/2ab

ae=1/2ac

de∥bcad=ae

ADE is an equilateral triangle.

ad=deade= adg+ edg=60° dfg is an equilateral triangle.

dg=dffdg=∠edg+∠edf=60°∠adg=∠edf

ADG EDF Congruence (SAS).

dag=∠def

dag=90°

de∥bc∠def=∠cfe

efc=90°

c=90°cef=180-90-60=30°

ce=1/2cf

ce^2=cf^2+ef^2

cf=1ef=√3

Solution: Connect DE

In equilateral ABC, ab=ac, c= bac=60° point d, e are the midpoints of ab and ac, respectively.

ad=1 2ab ae=1 2ac, ad=aede is the abc median, de bc def= cfe ade is an equilateral triangle (there is an isosceles triangle with an angle of 60° is an equilateral triangle) ad=de, ade= adg + edg=60° dfg is an equilateral triangle, dg=df

fdg=∠edg+∠edf=60°

adg=∠edf

adg≌△edf（sas)

dag=∠def

dag=90°∴∠efc=90°

In RT EFC, C=60°Tan60°=EF FCCF=1 EF= 3, got it.

ad=ae, then adc= aeb

and bad= eac, then bae= dac's two corners side theorem, we can get bae is all equal to cad ab=ac

Question 2, ad=ae, then ade= aed, so adb= ae and ad=ae, bad= cae

Hence the Bad Cae (ASA).

Hence ab=ac

Question 3, af de, so afc= edc, caf= ced180°= edc+ b= afc+ afb, so afb= b

Therefore ab=af=ed

Hence AFC EDC (ASA).

Therefore cf=cd