As shown in the figure, in the triangle ABC, ∠ 1 = ∠ 2, G is the midpoint of AD, lengthening BG intersecting AC with E.F is a point on AB, CF ⊥ ad is in H (1) Ad is the angular bisector of the triangle Abe; (2) Be is the center line on the side ad of the triangle abd; (3) Ch is the height on the edge ad of the triangle ACD A. 1 B. 2 C. Three D. 0

As shown in the figure, in the triangle ABC, ∠ 1 = ∠ 2, G is the midpoint of AD, lengthening BG intersecting AC with E.F is a point on AB, CF ⊥ ad is in H (1) Ad is the angular bisector of the triangle Abe; (2) Be is the center line on the side ad of the triangle abd; (3) Ch is the height on the edge ad of the triangle ACD A. 1 B. 2 C. Three D. 0

① The angle of ABC is the bisector of the angle;
② According to the concept of the middle line of the triangle, BG is known to be the middle line on the abd side ad of the triangle, so this option is wrong;
③ According to the concept of the height of the triangle, this option is correct
Therefore, a

As shown in the figure, in △ ABC, O is the intersection point of high AD and be. Observe the graph and try to guess what kind of quantitative relationship exists between ∠ C and ∠ doe, and prove your conjecture conclusion

∠C+∠DOE=180°．
∵ ad, be is the height of ∵ ABC (known),
Ψ AEO = ∠ ADC = 90 ° (high meaning),
∵ DOE is the external angle of  AOE (the concept of triangle external angle),
Ψ DOE = ∠ OAE + ∠ AEO (one outer angle of a triangle is equal to the sum of two nonadjacent interior angles)
=∠OAE+90°（∠AEO=90°）
=∠OAE+∠ADC（∠ADC=90°）
∴∠C+∠DOE=∠OAE+∠C+∠ADC=90°+90°=180°．
Another method: in the quadrilateral ceod, ∠ C + ∠ EOD + 90 ° + 90 ° = 360 °,
Then ∠ C + ∠ EOD = 180 °

As shown in the figure, in △ ABC, O is the intersection point of high AD and be. Observe the graph and try to guess what kind of quantitative relationship exists between ∠ C and ∠ doe, and prove your conjecture conclusion

∠C+∠DOE=180°．
∵ ad, be is the height of ∵ ABC (known),
Ψ AEO = ∠ ADC = 90 ° (high meaning),
∵ DOE is the external angle of  AOE (the concept of triangle external angle),
Ψ DOE = ∠ OAE + ∠ AEO (one outer angle of a triangle is equal to the sum of two nonadjacent interior angles)
=∠OAE+90°（∠AEO=90°）
=∠OAE+∠ADC（∠ADC=90°）
∴∠C+∠DOE=∠OAE+∠C+∠ADC=90°+90°=180°．
Another method: in the quadrilateral ceod, ∠ C + ∠ EOD + 90 ° + 90 ° = 360 °,
Then ∠ C + ∠ EOD = 180 °

As shown in the figure, it is known that in △ ABC, ad is high, CE is midline, DC = be, DG ⊥ CE, G is perpendicular foot It is proved that: (1) g is the midpoint of CE; (2) B = 2 ∠ BCE

It is proved that: (1) connect de; ∵ ad ⊥ BC, e is the middle point of AB, ᙽ De is the center line on the oblique side of RT ⊥ abd, namely de = be = 12ab; ∵ DC = de = be; and ∵ DG = DG,

As shown in the figure, it is known that in △ ABC, ad is high, CE is midline, DC = be, DG ⊥ CE, G is perpendicular foot It is proved that: (1) g is the midpoint of CE; (2) B = 2 ∠ BCE

It is proved that: (1) connect de; ∵ ad ⊥ BC, e is the middle point of AB, ᙽ De is the center line on the oblique side of RT ⊥ abd, namely de = be = 12ab; ∵ DC = de = be; and ∵ DG = DG,

As shown in the figure, it is known that in △ ABC, ad is high, CE is midline, DC = be, DG ⊥ CE, G is perpendicular foot It is proved that: (1) g is the midpoint of CE; (2) B = 2 ∠ BCE

It is proved that: (1) connect de; ∵ ad ⊥ BC, e is the middle point of AB, ᙽ De is the center line on the oblique side of RT ⊥ abd, namely de = be = 12ab; ∵ DC = de = be; and ∵ DG = DG,

As shown in the figure, in the triangle ABC, the angle bisectors ad, be and CF intersect at point h, passing through point h as Hg vertical AB, and the foot perpendicular to g, then is the angle AHF = BHG Why? I want to draw pictures, but I can't put them/~

From the opposite vertex angle AHF = CHD = 180 - (HCD + CDH) (1) and because CDH = DAB + DBA = DAB + 2hbg (2) brings (2) into (1), AHF = 180 - (HCD + DAB + 2hbg) (3) because the sum of inner angles of triangle is 180, half of the inner angles add up to 90, so HCD + DAB + HbG = 90 is brought into (3)

In △ ABC, ad be CF are three midlines that intersect at a point g. what is the relationship between △ AGF and the area of △ age? As the title

∵ D is the midpoint of BC
∴S△ABD=S△ACD,S△BDG=S△CDG
∴S△ABG=S△ACG
∵ F and E are the midpoint of AB and AC respectively
∴S△AFG=S△BFG,S△AEG=S△CEG
∴2S△AFG=2S△AEG
∴S△AFG=S△AEG

As shown in the figure, in △ ABC, ad, be and CF are three midlines, which intersect at the same point G, (1) the area of △ AGF and △ age

Since there is no point, I'll simplify it and give you an idea
Area abd = area ACD (base and height are equal)
Area GBD = area GCD
So we only need to prove that area FGB = area EGC
It can be obtained by the area FBC = ECB (they can be obtained by subtracting the overlapping part from the overlap part)
Answer: handsome brother and

As shown in the figure, points E and F are respectively on the sides BC and Ca of the equilateral triangle ABC, be = CF, AE and BF intersect with point G, and calculate the degree of ∠ AGF

ABC is an equilateral triangle
AB=BC,∠ABC=∠BCA=60°
BE=CF
△ABE≌△BCF
∠BAE=∠CBF
∠AGF=∠BGE=∠ABG+∠BAE
=∠ABG+∠CBF
=∠ABC=60°
∠AGF=60°