If line AB = 8cm, C is the midpoint of AB, D is on CB, DB = 1.5cm, then line CD=______ cm．

According to the concept of the midpoint of the line segment, BC = 12ab = 4, so CD = bc-bd = 4-1.5 = 2.5

It is known that if BD = 4cm, extend DB to a so that Ba = 5cm, and point C is the midpoint of line ad, then BC=

BD = 4, Ba = 5, extend DB to A. my understanding is that a and D are on both sides of B, then ad = 5 + 4 = 9, C is the midpoint of AD, that is, AC = 4.5, BC = AB-AC = 0.5. If you change the direction, BC is also 0.5

If the line segment AB is known, extend AB to C to make BC = AB, and then extend Ba to D to make ad = AC, then DB ∶ AB = (), CD ∶ BD = ()

DC=AD+AB+BC=3AB+AB+2AB=6AB=6cm
DB=DA+AB=3AB+AB=4AB=4cm
DA=3AB=3cm

Given the line segment AB, take a point C on the extension line of AB so that BC = 2Ab, and then take a point D on the extension line of Ba so that Da = AB and take the midpoint e of ab,
If de = 7.5cm, find the length of DC~

∵ Da = AB, point E is the midpoint of AB, de = 7.5cm
∴DE=DA+1/2AB=3/2AB
∴AB=5,
∵BC=2AB,
∴DC=DA+AB+BC
=5+5+10
=20cm

As shown in the figure, point C is the midpoint of line AB, point D is the midpoint of line BC, ad = 4cm, then what are the lengths of CD and ab?
┕━━━━━━┻━━━┻━━━━┙
A C D B

A——C—D—B
∵ C is the midpoint of ab
∴AC＝BC＝AB/2
∵ D is the midpoint of BC
∴CD＝BD＝BC/2＝AB/4
∴AD＝AC+CD＝AB/2+AB/4＝3AB/4
∵AD＝4
∴3AB/4＝4
∴AB＝16/3(cm)
∴CD＝AB/4＝4/3(cm)

Given that the line segments ad = 6cm, AC = 4cm, BD = 4cm, e and F are the midpoint of AB and CD respectively, how many centimeters is ef?
There should be analysis and process

Suppose the position of a is 0,
Because ad = 6cm, the position of D is 6;
Because AC = 4cm, the position of C is 4;
Because BD = 4cm, the position of B is 2;
E is the midpoint of AB, so the position of E is 1;
F is the midpoint of CD, so the position of F is 5;
So EF is 4cm
If you don't understand, draw a number axis and mark (0 ~ 6) 7 points
Mark a at 0, then check and mark other letters. It's very simple

As shown in the figure, two points B and C divide the line ad into three parts: 4:5:7. E is the midpoint of the line ad, CD = 14 cm
(1) (2) the value of AB: be

Let AB, BC and CD be 4x cm, 5x cm, 7x cm, respectively, ∵ CD = 7x = 14, ∵ x = 2. (1) ∵ AB = 4x = 8 (CM), BC = 5x = 10 (CM), ∵ ad = AB + BC + CD = 8 + 10 + 14 = 32 (CM), so EC = 12ad-cd = 12 × 32-14 = 2 (CM); (2) ∵ BC = 10 cm, EC = 2 cm, ∵ be = bc-ec = 10-2 = 8 cm, and ∵ AB = 8 cm, ∵ AB: be = 8:8 = 1 The value of AB: be is 1

Given that point C and point D are two points on line AB, and ab = ACM CD = BCM, point m is the midpoint of line AC, and point n is the midpoint of BD, the length a of line Mn is obtained D M_ C N________ B
Urgent help

I didn't calculate it carefully, but made a few assumptions. The result of reasoning is (a-b) / 2
If you simplify this condition, suppose that two points of CD coincide and are the midpoint of AB, so it's easy to calculate
Or, the midpoint of CD is on the midpoint of ab
Or point C coincides with point B, and point d coincides with point a
The conclusion of these hypotheses can be deduced that Mn is (a-b) / 2
(PS: maybe not)

As shown in the figure, C is the midpoint of line AB, D is the midpoint of line CB, CD = 1cm, find the length sum of AC + AD + AB in the figure

∵ CD = 1cm, D is the midpoint of CB, ∵ BC = 2cm, and ∵ C is the midpoint of AB, ∵ AC = 2cm, ab = 4cm, ∵ ad = AC + CD = 3cm, ∵ AC + AD + AB = 9cm

As shown in figure (1), point C is a point on line AB, triangle ACM and triangle BCN are equilateral triangles
(1) Verification: an = BM
(2) If De is connected, is there any new conclusion?
(3) If the original question "triangle ACM and triangle BCN are two equilateral triangles" is replaced by two squares (as shown in figure (2)), what is the relationship between an and BM? Please explain the reason
M N
O
D E
A C B
(1)
N E
F M
D
A B
C

Delta AMC is an equilateral triangle
∴AC = MC,∠ACM =60°
∵△ NCB is an equilateral triangle
∴MC = CB,∠NCB = 60°
∴∠ACN = ∠MCB
∴△ACN≌△MCB
∴AN = MB

If line AB = 8cm, C is the midpoint of AB, D is on CB, DB = 1.5cm, then line CD=______ cm．