As shown in the figure, △ ABC is an isosceles right triangle, ∠ ACB = 90 ° and ad is the center line on the side of BC. Through C, the vertical line of ad is made, which intersects AB at point E and ad at point F. it is proved that ∠ ADC = ∠ BDE

As shown in the figure, △ ABC is an isosceles right triangle, ∠ ACB = 90 ° and ad is the center line on the side of BC. Through C, the vertical line of ad is made, which intersects AB at point E and ad at point F. it is proved that ∠ ADC = ∠ BDE

Make ch ⊥ AB at h and ad at P,
∵ in RT △ ABC, AC = CB, ∵ ACB = 90 °,
∴∠CAB=∠CBA=45°．
∴∠HCB=90°-∠CBA=45°=∠CBA．
The midpoint of BC is d,
∴CD=BD．
And ∵ ch ⊥ ab,
∴CH=AH=BH．
And ? PAH + ∠ APH = 90 °, PCF + ∠ CPF = 90 °, APH = ∠ CPF,
∴∠PAH=∠ECH．
In △ APH and △ CEH
∠PAH=∠ECH，AH=CH，∠PHA=∠EHC，
∴△APH≌△CEH（ASA）．
∴PH=EH，
And ∵ PC = ch-ph, be = bh-he,
∴CP=EB．
∵ △ ACB is an isosceles right triangle,
∴∠B=45°，
That is ∠ EBD = 45 °,
∵CH⊥AB，
∴∠PCD=45°=∠EBD，
In △ PDC and △ EDB
PC=EB，∠PCD=∠EBD，DC=DB，
∴△PDC≌△EDB（SAS）．
∴∠ADC=∠BDE．

In the triangle ABC, a = 3, B = 4, C = 6. H (a) denotes the height on the edge of A. H (b), H (c) is similar to (HA + HB + HC) (1 / HA + 1 / HB + 1 / HC)

According to Helen's formula, the triangle area s = √ [P (P-A) (P-B) (P-C)] P = (a + B + C) / 2 = 13 / 2
Because 2S = a * ha = b * HB = C * HC, HA / HB = B / A, and so on
It is found that = 3S + (HB + HC) / HA + (HA + HC) / HB + (HA + HB) / HC
=3S+a/b+a/c+b/a+b/c+c/a+c/b=3s+27/4=105/4

Let a, B, C be the length of the three sides of the acute triangle ABC, ha, Hb, HC be the height of the corresponding side, then the value range of u = HA + HB + HC / A + B + C is U=（ ）A 0＜U＜1/2 B 1/2＜U＜1 C 1＜U＜2 D 2＜U＜4 Let a, C be the length of the three sides of the acute triangle ABC, ha, Hb, HC be the height of the corresponding side, then the value range of u = HA + HB + HC / A + B + C is u = () A 0＜U＜1/2 B 1/2＜U＜1 C 1＜U＜2 D 2＜U＜4

Analysis: first draw the graph according to the meaning of the title, then HA + BD > C, HA + DC > b, 2ha + a > b + C, the same reason, 2hb + b > C + A, 2hc + C > A + B, 2 (HA + HB + HC) > (a + B + C), and hA < B, Hb < C, HC < A, HA + HB + HC < A + B + C, and then the answer can be obtained
 
 
As shown in the figure below:
∵ha+BD＞c,ha+DC＞b,
∴2ha+a＞b+c,
Similarly, 2hb + b > C + A, 2hc + C > A + B,
∴2（ha+hb+hc）＞（a+b+c）,
Ha < B, Hb < C, HC < a,
∴ha+hb+hc＜a+b+c
∴U＜1
so
1/2＜U＜1．

In the right triangle ABC, ∠ C = 90 degrees, AC = 4, then vector AB times vector AC equals?

The original formula = | ab | * | AC | cosa = | ab | * (| AC | / | ab |) = 4 * 4 = 16

In the right triangle ABC, if the vector AB = (1, K), the vector AC = (2,1), and the angle c is equal to 90 degrees, then the value of K

Vector BC = ac-ab = (1,1-k),
The angle c is 90 degrees,
∴0=AC*BC=2+1-k=3-k,
∴k=3.

In the right triangle ABC, the angle a is 90 degrees and ab = 1. Find the value of vector AB dot multiplication vector BC

A rectangular coordinate system is established with a as the origin, AB as the x-axis and AC as the y-axis
So a (0,0) B (1,0) C (0, y)
So the vector AB = (1,0)
Vector BC = (- 1, y)
So vector AB dot product vector BC = 1 * (- 1) + 0 * y = - 1

In the right triangle ABC, the angle c = 90 degrees, ab = 5, AC = 4. Find the scalar product of vector AB and vector BC

|BC|=3,AB*BC=|AB||BC|cos(π-∠B)
=5*3*(-3/5)
= -9

Let K ∈ Z be known, AB=（k，1）， AC = (2,4), if| If ab | ≤ 4, then △ ABC is a right triangle___ .

From the meaning of the title
AB=（k，1），|
AB|≤4，
Therefore, there are K2 + 1 ≤ 16, and K ∈ Z. therefore, there are seven kinds of K, which may be - 3, - 2, - 1, 0, 1, 2, 3, that is, there are seven such triangles,
also
AC = (2,4), so vector
BC=（2-k，3），
order
AB•
AC = 0, 2K + 4 = 0, k = - 2,
order
AB•
BC = 0 leads to 2k-k2 + 3 = 0, k = - 3, or K = 1,
order
AC•
BC = 0, get 4-2k + 12 = 0, get k = 8,
So the number of right triangles is 3,
The probability that △ ABC is a right triangle is 3
7；
So the answer is: 3
7．

Given that the sum of the two right angles of a right triangle is equal to 8, what is the maximum area of the right triangle?

Let the right side of a right triangle be x, then the other right side is 8-x. the area of a right triangle is s
According to the meaning of the title, we get
S=1
2x（8-x）（0＜x＜8），
Formula, get
S=-1
2（x-4）2+8；
When x = 4, that is, when the two right angles are 4, the area of the triangle is the largest, and the maximum area is 8

Given that the sum of two right angles of a right triangle is equal to 4, what is the maximum area of the right triangle? Quadratic function

Let one right side of the right triangle be x and its area be y, then the other right angle side will be (4-x)
According to the triangle area formula, the quadratic function is obtained
Y = x (4-x) △ 2, that is, y = - X / / 2 + 2x
The result is as follows: 1
y=-1/2(x-2)²+2
So: when the two right sides of the triangle are 2, the area of the right triangle is the largest, and the maximum value is 2