As shown in the figure, in △ ABC, ab = AC, ∠ BAC = 120 ° and EF is the vertical bisector of ab. EF intersects BC at point F and ab at point E. verification: BF = 1 2FC．

As shown in the figure, in △ ABC, ab = AC, ∠ BAC = 120 ° and EF is the vertical bisector of ab. EF intersects BC at point F and ab at point E. verification: BF = 1 2FC．

Proof: connect AF,
∵AB=AC，∠BAC=120°，
∴∠B=∠C=30°，
∵ EF is the vertical bisector of ab,
∴BF=AF，
∴∠BAF=∠B=30°，
∴∠FAC=120°-30°=90°，
∵∠C=30°，
∴AF=1
2CF，
∵BF=AF，
∴BF=1
2FC．

In the triangle ABC, CD / DA = AE / EB = 1 / 2, vector BC = vector a, vector CA = vector B, proof vector de = 1 / 3 (vector b-vector a) Please give me an easy to understand and specific method. I can't learn fast

Vector AB = - (vector a + vector b) vector de = vector Da + vector AE vector Da = 2 / 3 vector CA = 2 / 3 vector b vector AE = 1 / 3 vector AB = - 1 / 3 (vector a + vector b), so vector de = 1 / 3 (vector b-vector a)
This is mainly about drawing pictures and finding relationships

In the triangle ABC, CD / DA = AE / EB = 1 / 2, note vector BC = a, vector CA = B, use a, B to represent vector De

AB=-a-b
AE=(-a-b)/3
DA=2b/3
DE=DA+AE=(b-a)/3

As shown in the figure, in △ ABC, D and F are the midpoint of BC and Ca respectively, vector AE = 2 | 3, vector AC, vector AB = a, vector AC = B （1） A B for vector AD AE BE BF (2) Prove B e f collinear

Vector ad = (vector a + vector b) / 2 vector AE = two thirds vector ad = (vector a + vector b) / 3 vector AF = vector AC / 2 = vector B / 2 vector BF = vector Ba + vector AF = - vector a + vector B / 2 vector be = vector Ba + vector AE = - vector a + (vector a + vector b) / 3 = (- 2 vector a + vector b) / 3 2 vector BF = to

In the triangle ABC, D, e and F are the midpoint of BC, Ca and ab respectively. It is proved that vector Da + vector EB + vector FC = vector 0

Vector Da + vector EB + vector FC
=Vector DC + vector Ca + vector EA + vector AB + vector FB + vector BC
=Vector DC + vector BC + vector EA + vector Ca + vector FB + vector ab
=3 / 2 * (vector BC + vector Ca + vector AB)
=Vector 0

As shown in the figure, in the RT triangle ABC, ∠ ACB = 90 ° D is the midpoint of AB, de ⊥ DF, if, ca

Let d be DG, so that DG = BD, angle BDF = angle FDG, and connect eg to triangle BDF and triangle GDF, and get two triangles congruent, BF = FG, angle c = angle DGF to triangle ade and triangle GDE, because ad = BD = DG, de = De, angle GDE = 90 degrees - angle GDF = 180 degrees - angle EDF - angle BDF = angle ade

In the triangle ABC, D is the midpoint of ab. extend Ca, CB to e, f respectively, so that DF = De, passing through e, F as the vertical line of Ca and CB, intersecting at point P, find ∠ PAE = ∠ PBF

Take point m of PA and point n of Pb
Because m and N are the mid points of RT △ AEP and RT △ BFP respectively,
Therefore, EM = am, FN = BN
Because DM and DN are the median lines of △ PAB
So DM ‖ BN, DM = BN, DN ‖ am, DN = am
And DM = BN = NP = NF, DN = am = MP = me
And ∠ amd = ∠ Bnd = ∠ APB
Because de = DF, so △ DEM ≌ △ FDN
If the corresponding angles are equal, then
∠EMD＝∠FND
Then, ame = BNF
And △ ame and △ BNF are isosceles triangle
Therefore, PAE = PBF

It is known that in △ ABC, CB = Ca, angle c = 90 degrees, D is any point of AB, AE is perpendicular to CD, perpendicular foot is e, BF is perpendicular to CD, vertical angle is 90 degrees Additional: I don't know Pythagorean theorem! It is known that in △ ABC, CB = Ca, angle c = 90 degrees, D is any point of AB, AE is perpendicular to CD, perpendicular foot is e, BF is perpendicular to CD, perpendicular foot is f Verification: EF = / ae-bf/

It is proved that: (in the case of AE ⊥ BF, i.e. D is close to point B, the other case is the same method) because AE ⊥ CD, ﹣ EAC ≌ ace ≌ BCF ≌ ACB = 90 ≌ so: ≌ BCF ≌ ace ≌△ ace

It is known that in △ ABC, CA = CB, ∠ C = 90 ° D is any point on AB, AE ⊥ CD, perpendicular foot is e, BF ⊥ CD, and vertical foot is F. it is proved that EF = | ae-bf |

It is proved that: ? AE ⊥ CD,  AEC = 90 degrees,  ACE + ∠ CAE = 90 degrees, (two acute angles of a right triangle are complementary) ? ? ACE +  BCF = 90 degrees, ? CAE = ∵ BCF, (the remainder of equal angle is equal) ? AE ⊥ CD, BF ⊥ CD, ? AEC = ∠ BFC = 90 °

It is known that: in the triangle ABC, CA = CB, angle c = 90 °, D is any point of AB, AE ⊥ CD, perpendicular foot is e, BF ⊥ CD, perpendicular foot is F. verification: BF = ae-bf

AC = CB
∠AEC=∠BFC=90°
∠CAE=∠BCF
The results show that △ ace is equal to △ CBF
∴AE=CF,BF=CE
AE-BF=CF-CE=EF
Your proof is wrong