Let D, e, f be the midpoint of BC, AC and ab of the triangle ABC respectively How to do such a topic,

Let D, e, f be the midpoint of BC, AC and ab of the triangle ABC respectively How to do such a topic,

Generally, according to the triangle rule after drawing, i.e. vector AB + BC = AC, because BC takes the key point of AB as the starting point. For AD, ad can be extended to the same length. According to the parallelogram rule, ad = (AB + AC) / 2be = (Ba + BC) / 2cf = (Ca + CB) / 2 are added to get AD + be + CF = 0.5 (AB + AC + BA)

In the triangle ABC, D is the midpoint of BC, e, f is the midpoint of AC, Ba, ad, be, CF intersect with O. vector Proof vector OA + ob + OC = 0

Extend Ao to o 'so that Ao = OO'
Then o, F and E are the midpoint of Ao ', AB and AC respectively
Of and OE are the median lines of △ ABO 'and △ ACO', respectively
∥ of ∥ o'b, OE ∥ o'c, i.e., Co ∥ o'b, Bo ∥ o'c
The quadrilateral BOCO 'is a parallelogram
ν OO '= ob + OC, and AO = OO'
∴ OA+OB+OC = OA+OO' = OA+AO = 0

In the triangle ABC, if ad is the midline on the edge of BC, be is the midline on AC and be is equal to a vector and ad is equal to B vector, then BC vector is

Vector a = be = (1 / 2) (Ba + BC),
Vector b = ad = (1 / 2) (AB + AC) = (1 / 2) (2Ab + BC),
∴2a+b=(3/2)BC,
The vector BC = (4 / 3) a + (2 / 3) B

In the triangle ABC, the vector ad = 1 / 4, vector AB, de ‖ BC, intersects with edge AC at point E, and the center line am and De of triangle ABC intersect at point n, Let vector AB = a, vector AC = B, try to use a, B for vector AE, vector BC, vector dB, vector EC, vector DN, vector an

In this paper, we make a point of the BC and de 124124\\\\\\\\\124\\\\124\\\124\124124\124\124\124\124\124a + 1 / 8b-1 / 8A = 1 / 8A + 1 / 8b

In the triangle ABC, ab = 2. AC = 3. Point D is the center of gravity of triangle ABC, then what is the vector ad multiplied by vector BC?

Since D is the center of gravity, ad = 1 / 3 * (AB + AC),
BC = ac-ab,
So ad * BC = 1 / 3 * (AB + AC) * (ac-ab) = 1 / 3 * (AC ^ 2-AB ^ 2) = 1 / 3 * (9-4) = 5 / 3

It is known that in the isosceles right triangle ABC, the angle ACB = 90 degrees, D is the midpoint of BC, CE is perpendicular to AD and intersects AB with e. it is proved that the angle CDF = angle BDE

Proof 1: cross C as CF ⊥ AB, respectively cross ad, AB in G, F. ∵ isosceles right triangle ABC, ? ACB = 90 degrees, ∵ AC = BC, and ∠ cab ⊥ B = 45 degrees. ? CF ⊥ AB, ∵ cab ⊥ B = 45 degrees, ∵ ACG ⊥ DCG = 45 °. AC ⊥ CD, CF ⊥ ad, ∵ CAG ∵ BCE

As shown in the figure △ ABC and △ CDE are equilateral triangles

It is proved that ∵ △ ABC and △ CDE are equilateral triangles,
∴BC=AC，CE=CD，∠BCA=∠ECD=60°，
∴∠BCA+∠ACE=∠ECD+∠ACE，
That is ∠ BCE = ∠ ACD,
∵ in ᙽ BCE and △ ACD,
BC＝AC
∠BCE＝∠ACD
CE＝CD ，
∴△BCE≌△ACD（SAS），
∴AD=BE．

As shown in the figure, in the triangle ABC, point D is the midpoint of BC. Make ray ad, take e and F on the line segment AD and the extension line respectively to connect CE and BF, and get the triangle BDF congruent CDE

The conditions are: DF = de (or CE ‖ BF or ∠ ECD = ∠ DBF or ∠ Dec = ∠ DFB, etc.)
The reasons are as follows.
∵ point D is the midpoint of BC,
∴BD=CD．
In △ BDF and △ CDE,
A kind of
BD＝CD
∠BDF＝∠CDE
DF＝DE
,
∴△BDF≌△CDE（SAS）．
So the answer can be: DF = De

As shown in the figure, in the isosceles right angle △ ABC, ∠ ACB = 90 ° D is the midpoint of BC, CE ⊥ ad intersects AB at e at F, and proves: ∠ CDF = ∠ BDE

It is proved that if the extension of crossing point B is the vertical line of BC, then the extended line of intersection CE is at m.cb = Ca, ∠ ACB = 90 °, then ∠ CBE = 45 ° = ∠ MBE. If CE is perpendicular to ad, then ∠ CDA = ∠ CMB (both are the remainder angles of angle DCF); and Ca = CB; ∠ ACD = ∠ CBM = 90 °, then ⊿ ACD ≌⊿ CBM (AAS), then BM = BD, ≌⊿ CMB

As shown in the figure, in the isosceles triangle ABC, ABC = 90 ° AC = BC, D is the midpoint of BC, CE ad, and the foot perpendicular is F. try to explain that CDF = BDE

As shown in the figure, in the isosceles triangle ABC, ∠ ABC = 90 °, AC = BC, D is the midpoint of BC, CE ⊥ ad, perpendicular foot is f, try to explain ∠ CDF = ∠ BDE It is found that CN ⊥ AB intersects AB to N, ad to m, ∵ AC = BC, ∵ cam = ∠ BCE