As shown in the figure, given that D is a point on the extension line of edge BC of △ ABC, DF ⊥ AB is at point F, intersects AC at point E, ∠ a = 40 ° and ∠ d = 30 °, then the degree of ∠ ACB is obtained______ Degree

As shown in the figure, given that D is a point on the extension line of edge BC of △ ABC, DF ⊥ AB is at point F, intersects AC at point E, ∠ a = 40 ° and ∠ d = 30 °, then the degree of ∠ ACB is obtained______ Degree

In △ DFB,
∵DF⊥AB，
∴∠DFB=90°，
∵∠D=30°，∠DFB+∠D+∠B=180°，
∴∠B=60°．
In △ ABC,
∠A=40°，∠B=60°，
∴∠ACB=180°-∠A-∠B=80°．
So the degree of ∠ ACB is 80 degrees

As shown in the figure: there is a point D on the extension line of the edge ab of △ ABC. The crossing point D is DF ⊥ AC at f and BC at e, and BD = be. It is proved that △ ABC is an isosceles triangle

Proof: ∵ DF ⊥ AC,
∴∠DFA=∠EFC=90°．
∴∠A=∠DFA-∠D，∠C=∠EFC-∠CEF，
∵BD=BE，
∴∠BED=∠D．
∵∠BED=∠CEF，
∴∠D=∠CEF．
∴∠A=∠C．
The △ ABC is an isosceles triangle

As shown in the figure, in the triangle ABC, ab = AC, the straight line DF intersects AB ` BC at d ` e, the extension line of intersection AC is at F, and norbd = CF, is e the midpoint of DF Pictures can't be sent

yes
Proof: DG ∥ AC is given to BC to g
Then ∠ DGB = ∠ ACB = ∠ ABC
∴DG=DB=CF
DGC = ∠ ECF, ∠ DEG = ∠ CEF
∴△DGE≌△FCE
So de = EF
That is, e is the midpoint of DF

As shown in the figure, in △ ABC, ab = AC, the straight line DF intersects AB at D, the extension of AC is at point F, BC is at point E. if BD = CF, can you prove that e is the midpoint of DF? (sorry Sorry, please draw your own picture

Extend BC, make a parallel line of AB through point F and intersect the extension line of BC at point G
∵AB‖FG
∴∠B=∠G
∵AB=AC
∴∠B=∠ACB
∴∠ACB=∠G=∠FCG
∴FC=FG=BD
And ∵ bed = ∵ gef
∴△BDE≌△GFE
∴DE=EF
/ / E is the midpoint of DF

As shown in the figure, ad is the angular bisector of the triangle ABC, where de / / AB intersects AC at point E, DF / / AC crosses AB at point F, and extension of Fe intersects extension of BC at point G Confirmation: Ag = DG Angle GAC = angle B

I'm a math teacher. You draw your own pictures,
It is proved that: (1) ∵ ad bisection ∠ ABC
∴∠BAD=∠CAD
∵DE//AB ,DF//AC
∴∠BAD=∠ADE,∠CAD=∠FDA
∵∠BAD=∠CAD
Therefore, AF = DF
Afde is a parallelogram
The quadrilateral afde is a diamond
ν Fe vertically bisects ad
/ / FG is the vertical bisector of AD
∵ the distance from the point on the vertical bisector to both ends of the line segment is equal
∴AG=DG
（2）∵AG=DG
The △ ADG is an isosceles triangle
∴∠AGE = ∠DGE
∴△AEG≌△DEG
∴∠GAC = ∠GDE
∵DE//AB
∴∠GDE = ∠B
∴∠GAC = ∠B

As shown in the figure, AB is parallel to DC, ∠ ABC = ∠ ADC, AE = CF, be = DF. It is proved that EF and AC are equally divided

Connect AF, CE,
∵AB∥DC,∠ABC=∠CDA,
Then we put △ ACD around the midpoint of AC
After 180 ° rotation, it can be completed with △ Abe
Total weight
 AB = CD, ∠ BAC = ∠ DCA,
In △ Abe and △ CDF,
∵AE= CF,BE=DF,AB=CD,
Then △ CDF rotates around the midpoint of EF
After 180 ° rotation, it can be completely connected with △ Abe
Coincidence,
∴∠EAB=∠FCD,∴∠EAC=∠EAB
+∠BAC=∠FCD
+∠DCA=∠ACF,∴AE∥FC,
AE = FC,  quadrilateral aecf
The diagonal is a parallelogram
Share equally with AC

AB parallel CD, angle ABC = angle ADC, AE = CF

Now we can conclude that ABCD is a parallelogram,
However, it is impossible to prove that AE is parallel to CF, which should be a lack of conditions

AE is the center line of the triangle ABC, de bisection ∠ BDA intersects AB at e, DF bisection ∠ ADC intersects AC at F, and verification: be + CF > EF

Title: Ad (here d) is the ABC midline of triangle, de bisection ∠ BDA intersects AB in E, DF bisection ∠ ADC intersects AC in F, verification: be + CF > EF
Proof: make CM ∥ AB through C, turn ED extension line at point M, connect to FM
So ∠ B = ∠ DCM, ∠ bed = ∠ CMD,
And AE is the center line of the triangle ABC,
So BD = CD,
So △ BDE ≌ △ CDM
So be = cm, ed = MD
Because of the fact that BDE ∠,
So ∠ ade = ∠ ADB / 2,
Because DF bisection ∠ ADC intersects AC with F,
So ∠ ADF = ∠ ADC / 2,
So ∠ ade + ∠ ADF = ∠ ADB / 2 + ∠ ADC / 2 = (∠ ADB + ∠ ADC) / 2,
Because ∠ ADB + ∠ ADC = 180,
So ∠ ade + ∠ ADF = 90 °,
So FD divides EM vertically,
So EF = FM,
In triangular CFM, CM + FC > FM,
Be + CF > EF

As shown in the figure: in △ ABC, ab = AC, D is any point on BC, de ∥ AC intersects AB at e, DF ∥ AB intersects AC at F. verification: de + DF = AC

Proof: ∵ DE ∵ AC, DF ∵ AB,
The quadrilateral AEDF is a parallelogram,
∴DE=AF，
AB = AC,
∴∠B=∠C，
∵DF∥AB，
∴∠CDF=∠B，
∴∠CDF=∠C，
∴DF=CF，
∴AC=AF+FC=DE+DF．

As shown in the figure: in △ ABC, ad is its angular bisector. It is proved that s △ abd: s △ ACD = AB: AC

It is proved that: de ⊥ AB, DF ⊥ AC, the vertical foot is e, F,
∵ ad bisection ∵ BAC,
∴DE=DF，
∴S△ABD：S△ACD=（1
2×AB×DE）：（1
2×AC×DF），
=AB：AC．