As shown in the figure, the straight line DF intersects with AB and AC on both sides of △ ABC at two points D and e respectively, and intersects with the extension line of BC at point F, ∠ B = 50 °, 1 = 76 °, f = 3

As shown in the figure, the straight line DF intersects with AB and AC on both sides of △ ABC at two points D and e respectively, and intersects with the extension line of BC at point F, ∠ B = 50 °, 1 = 76 °, f = 3

∵ 1 = 76 ° (known)
﹤ ade = 76 ° (equal vertex angle)
∵ B + ∠ f = ∠ ade (an outer angle of a triangle is equal to the sum of two interior angles not adjacent to it)
Ψ ade = 50 ° + 30 ° = 80 ° (lighting property)
∵ a + ∠ ade + ∠ AED = 180 ° (the sum of inner angles of triangle is 180 °)
Ψ a = 180 ° - ∠ AED - ∠ ade = 24 ° (equality property)

As shown in the figure, AB bisects CD, ∠ ABC = ∠ ADC, AE = CF, be = DF

What about the picture?

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，
AC = ab,
∴∠B=∠C，
∵DF∥AB，
∴∠CDF=∠B，
∴∠CDF=∠C，
∴DF=CF，
∴AC=AF+FC=DE+DF．

In △ ABC, if ad is an angular bisector of ∠ BAC, point E and point F are on AB and AC respectively, and de ⊥ AB, the perpendicular foot is e, DF ⊥ AC, and the perpendicular foot is f (as shown in Fig. (1)), the following two conclusions can be obtained: ①∠AED+∠AFD=180°；②DE=DF． Then, in △ ABC, there is still the condition that "ad is the angular bisector of ∠ BAC, and point E and point F are on AB and AC respectively", please explore the following two questions: (1) If ∠ AED + ∠ AFD = 180 ° (as shown in Fig. (2)), are de and DF still equal? If not, please give an example (2) If de = DF, is ∠ AED + ∠ AFD = 180 ° true? (only write the conclusion, not prove it)

(1) The reasons are as follows: the crossing point D is DM ⊥ AB in M, DN ⊥ AC in N, ∵ ad bisection ∵ BAC, DM ⊥ AB, DN ⊥ AC, ∵ DM = DN, ? AED + ∠ AFD = 180 °, AFD + ∠ DFN = 180 degrees,  DFN = AED, ⊥ DME ≌ △ DNF (AAS), ∵ de = DF; (2) does not necessarily hold

As shown in the figure, points a, B, D and E are on the same straight line, ad = EB, BC ‖ DF, ∠ C = ∠ F. verification: AC = EF

It is proved that: ∵ ad = EB ᙽ ad-bd = eb-bd, that is, ab = ed ? BC ∥ DF, ? CBD = ∠ FDB  ABC = ∠ EDF  in △ ABC and △ EDF, ? C ? f ? ABC = EDFA

As shown in the figure, points a, B, D and E are on the same straight line, ad = EB, BC ‖ DF, ∠ C = ∠ F. verification: AC = EF

It is proved that: ∵ ad = EB ᙽ ad-bd = eb-bd, that is, ab = ed ? BC ∥ DF, ? CBD = ∠ FDB  ABC = ∠ EDF  in △ ABC and △ EDF, ? C ? f ? ABC = EDFA

As shown in the figure, points a, B, D and E are on the same straight line, ad = EB, BC ‖ DF, ∠ C = ∠ F. verification: AC = EF

It is proved that: ∵ ad = EB ᙽ ad-bd = eb-bd, that is, ab = ed ? BC ∥ DF, ? CBD = ∠ FDB  ABC = ∠ EDF  in △ ABC and △ EDF, ? C ? f ? ABC = EDFA

As shown in the figure, points a, B, D and E are on the same straight line, ad = EB, BC ‖ DF, ∠ C = ∠ F. verification: AC = EF

Proof: ad = EB
/ / ad-bd = eb-bd, that is, ab = ed
And ∵ BC ∥ DF,
∴∠CBD=∠FDB     
∴∠ABC=∠EDF 
In △ ABC and △ EDF,
A kind of
∠C＝∠F
∠ABC＝∠EDF
AB＝ED
∴△ABC≌△EDF，
∴AC=EF

As shown in the figure, in △ ABC, D is a point on AC, e is a point on CB extension line, and AC BC＝EF DF， Confirmation: ad = EB

It is proved that DH ∥ BC is made through point D and ab is crossed with H, as shown in Fig,
∵DH∥BC，
∴△AHD∽△ABC，
∴AD
AC=DH
BC is ad
DH=AC
BC，
∵DH∥BE，
∴△BEF∽△HDF，
∴BE
HD=EF
DF，
And AC
BC＝EF
DF，
∴BE
HD=AD
DH，
∴AD=EB．

As shown in the figure, in the triangle ABC, ab = AC, take a point D on AB, and take a point E on the extension line of AC, so that BD = CE connects de and meets BC at point F. verification: DF = EF Quick answer

In this paper, we make the DG