In △ ABC, it is known that BD and CF are high, M is the midpoint of BC, and N is the midpoint of DF Yes,

prove:
Connect MD, MF
∵ BFC = 90 ° and M is the midpoint of BC
/ / FM = 1 / 2BC (the center line of the hypotenuse of a right triangle is equal to half of the hypotenuse)
Similarly, MD = 1 / 2BC
∴FM =DM
∵ n is the midpoint of DF
⊥ FD

In the triangle ABC, D is the midpoint of BC, and a straight line L and AB, AC intersect m and N respectively, and the vector am = XAB, vector an = YAC, Find the function analytic formula of Y with respect to X

Although the title has not been explained, I feel that the straight line L is the connection DM and DN that can intersect the extension line of AB and AC, and the midpoint of ad is oad = (AB + AC) / 2, DM = am-ad, DN = an-ad

In the triangle ABC, D is the midpoint of BC side. Vector am = m vector AB, vector an = n vector AC, Mn and ad intersect point P, vector AP = x vector AP (1) When m = 1, n = 0.5, find the value of X; (2) When m, n belong to (0,1), let m, n denote X

Condition: vector AP = x vector AP, should be ap = x vector ad (1) when m = 1, n = 0.5, am = AB, n is the midpoint of AC, so p is the center of gravity of triangle ABC, AP = (2 / 3) ad, that is, x = 2 / 3 (2) AB = (1 / M) am, AC = (1 / N) an, ad = (1 / x) AP and ab + AC = 2ad, so (1 / M) am + (1 / N) an = (2 / x) APAP = [x / (2m)] am

Ad is the middle line of the triangle ABC, am is vertical to AB, and am = AB, an is vertical to AC, an = AC, Mn = 2ad

As shown in the figure: make point e so that be is parallel and equal to AC, CE is parallel and equal to AB, then ABEC is a parallelogram,
∵AM=AB AN=AC AC=BE
∴AN=BE
And ∵ am ⊥ ab ⊥ an ⊥ AC ⊥ ABEC is a parallelogram
∴∠MAN=180°-∠BAC ∠ABE=180°-∠BAC
∴∠MAN=∠ABE
∴△MAN≌△ABE
∴MN=AE
Then ∵ AE = 2ad
So Mn = 2ad

It is proved that AF is perpendicular to the bisector AF, AF = be

Then connect HF ∵ EF ⊥ BC, Ag / / EC, ᚉ Ag ⊥ BC, and ab = AF, so AG is the perpendicular line of the triangle ABF.  BH = FH,  HBC = ∠ HFB, and EF is the vertical bisector of BC, be = CE,  ECB = ≁ HBC,  ECB = ∠ HFB, ᚉ HF / / AC, namely AE / / HF

In the triangle ABC, be bisection angle ABC be is perpendicular to F D is the midpoint of AB to prove DF parallel BC

Proof tips:
Extend ad to BC or BC extension line to g
It is easy to prove that f is the midpoint of Ag
(AD = DG for vertical, angular bisector, public boundary card, etc.)
Because D is the midpoint of ab
So DF is the median of the triangle ABG
So DF / / BC

In the triangle ABC, ab = AC, the central line BD, CE intersect with m, eg parallel BD, DF parallel CE, eg, DF compared with point n, it is proved that Mn bisects de vertically I'm sorry, I have to draw it myself. Sorry, sorry^_^

Let ah be the center line on BC, and it must pass through M. turn ABC to ah by 180 degrees. B and C coincide. E and D coincide. F and G coincide. N is on ah. Mn bisects ed vertically

In this paper, we prove that a B and a B are the midpoint of a graph

Proof: connect ad,
∵ AB = AC, D is the midpoint of BC,
∴∠EAD=∠FAD，
In △ AED and △ AFD,
AE＝AF
∠EAD＝∠FAD
AD＝AD ，
∴△AED≌△AFD（SAS），
∴DE=DF．

In triangle ABC, if vector BD = 3 vector DC, vector ad = m vector AB + n vector AC, then the value of Mn is

AD=AB＋BD
=AB＋(3/4)BC
=AB＋(3/4)(AC－AB)
=(1/4)AB＋(3/4)AC
Then: M = 1 / 4, n = 3 / 4
Then mn=3/16

It is known that the changes of triangle ABC are ABC, vector M = (2cos square a divided by 2,1), vector n = (3, cos2a), vector Mn = 4 (1) Find the size of angle a (2) If B-C = 1, a = 3, find the area of triangle ABC

（1）m*n=6[cos(A/2)]^2+cos(2A)
=3*(1+cosA)+2(cosA)^2-1=4 ,
(cosa-2) = 0,
Therefore, if cosa = - 2 (round off) or cosa = 1 / 2, then a = π / 3
(2) According to the cosine theorem, a ^ 2 = B ^ 2 + C ^ 2-2bccosa is obtained,
So B ^ 2 + C ^ 2-bc = 9, and B-C = 1,
So BC = (b ^ 2 + C ^ 2-bc) - (B-C) ^ 2 = 8,
Then SABC = 1 / 2 * BC * cosa = 1 / 2 * 8 * 1 / 2 = 2

In △ ABC, it is known that BD and CF are high, M is the midpoint of BC, and N is the midpoint of DF Yes,