Point C is the point on line AB, and point D is the midpoint of BC. If ad = 5cm, then AC + AB is

Point C is the point on line AB, and point D is the midpoint of BC. If ad = 5cm, then AC + AB is

AC + AB = 2 (AC + CD) = 2ad, so the final value is 10cm

As shown in the figure, given that M is the midpoint of AD, am = 5cm, B is the midpoint of AC, CD = 2 / 3aC, find the length of AB, BC and CD

AD=2×5=10(cm)

As shown in the figure, given that the line AB = 5cm, the point C is on AB, BC = 2 / 3aC, and the point D is the midpoint of BC, find the length of BC and AD

A——C——D——B
∵AB＝AC+BC,BC＝2AC/3,AB＝5
∴AC+2AC/3＝5
∴AC＝3
∴BC＝AB-AC＝5-3＝2（cm）
∵ D is the midpoint of BC
∴CD＝BC/2＝2/2＝1
∴AD＝AC+CD＝3+1＝4（cm）

The positive integer solution of bivariate linear equation 3x + y = 9 is______ ．

When x = 1, 3 × 1 + y = 9, the solution is y = 6; when x = 2, 3 × 2 + y = 9, the solution is y = 3; when x = 3, 3 × 3 + y = 9, the solution is y = 0 (not conforming to the meaning of the problem, rounding off). Therefore, the positive integer solution of the equation is x = 1y = 6, x = 2Y = 3. So the answer is: x = 1y = 6, x = 2Y = 3

If the solution of the system of linear equations {x + y = m, 3x + 4Y = 36 with respect to x, y is a positive integer, then M is an integer=

X plus y equals M
X equals m minus y
Substituting 3x and 4Y is 36
3 m plus y equals 36
Y equals 36 minus 3M
X equals 4m minus 36
Because x, y are positive integers
36 minus 3M greater than 0
4m minus 36 greater than 0
So m is greater than 9 and less than 12
M equals 10,11

Finding the non positive integer solution of the quadratic equation 3x = 4Y = - 20

You're wrong about this equation. How can you get two equal signs? If you're right, there's no process. X = - 20 / 3, y = - 5

In the triangular pyramid v-abc, VO ⊥ plane ABC, O ∈ CD, VA = VB, ad = BD. it is proved that CD ⊥ AB and AC = BC

It is proved that: VA = VB, ad = BD {VD ⊥ AB, VO ⊥ plane ABC, ab ⊂ plane ABC {VO ⊥ AB} plane VCD, CD ⊂ plane VCD {ab ⊥ CD, that is, CD ⊥ AB and ad = BD, CD = CD, ∠ BDC = ∠ ADC = 90 ° {ADC ≌ BDC} AC = BC

As shown in the figure, in the triangular pyramid v-abc, VO ⊥ plane ABC, O ∈ CD, VA = VB = 32, ad = BD = 3, BC = 5. (1) prove: VC ⊥ AB; (2) when the dihedral angle ∠ VDC = 60 °, find the volume of the triangular pyramid v-abc

(1) It is proved that connecting VD, ∩ VO = V, ∩ VC ⊥ AB, and VD ∩ VO = V, ∩ VC ⊥ AB are the middle points of AB, ∩ VA = VB = 32, ≁ VD = 3, in RT △ VDO, ∩ VD = 3, ∩ VC ⊥ ab

As shown in the figure, in the three-dimensional figure v-abc, VO ⊥ plane ABC, O on CD, VA = VB, ad = BD, can you determine Cd ⊥ AB and AC = BC?

Connecting VD, D is the midpoint of the isosceles triangle VAB, so AB is perpendicular to VD,
VO is perpendicular to plane ABC, AB is perpendicular to VO, so AB is perpendicular to plane VOD, so AB is perpendicular to CD
In triangle cab, D is the midpoint of AB, AB is perpendicular to CD, so triangle cab is isosceles triangle, so AC = BC

As shown in the figure, in the triangular pyramid v-abc, the bottom △ ABC is an equilateral triangle, VA = VB = VC = AB, VO ⊥ the bottom ABC is o,
M is the midpoint of VO, connecting Ma, MB and MC
Verification: Ma ⊥ plane MBC

First question:
Connect to OC
Then OC = √ 3bC / 3 = √ 3vc / 3
So VO = √ 6BC / 3, OM = √ 6BC / 6
MC=√(OC^2+OM^2)=√2BC/2
Similarly, MB = MC = √ 2BC / 2
So MB ^ 2 + MC ^ 2 = BC ^ 2
So MB ⊥ MC
Similarly, Ma ⊥ MC, Ma ⊥ MB
So Ma ⊥ plane MBC