There are four points of ABCD on the same straight line, AD 4 5DB, AC 9 5CB, and CD 4cm, find the AB

cd = 4 cm, so cd does not coincide with the celebration of the potato. So, point C and point D are between AB and point D and point BA extension. (otherwise coincide) let the chain between the positive ab is, so da=(5 4)aba=ac+bc=ac+(9 5)ac=(14 5)ac, so the honorable ac=(5 14)abcd=da+ac=(5 4)ab+(5 14)ab=(45...).

ac=ad+cd （1）

db=dc+cb （2）

ad=5 9db, ac=9 5cb, so db=9 5ad(1)+(2).

9/5（bc+ad）=ad+2cd+cb

4/5(bc+ad)=2cd

bc+ad=10

So ab=bc+ad+cd 14

I don't know if it's a coincidence, if you look annoyed, you can try ad=x, cb=y, find x+y, or if possible, find the value of x,y respectively.

by ad=5 9db, ad+db=ab, i.e. 5 9db +db=ab gets db=9 14ab by ac=9 5cb

ac+cb =ab, i.e., 9 5cb+ cb =ab, cb=5, 14 abcd= db - cb = 9, 14ab-5, 14 ab =4, i.e., 4 14ab=4

ab=14cm

If you don't understand, you can ask, and you can ensure that it is correct.

ac=ad+cd

db=dc+cb

ad=5 9db, ac=9 5cb, so db=9 5ad9 5(messy bc+ad)=ad+2cd+cb4 5(bc+ad)=2cd

bc+ad=10

So ab=bc+ad+cd 14

I don't know if it's a coincidence, if you look annoying, you can try to accompany Wang Jiezi ad=x,cb=y,find x+y or if possible, find the value of x,y respectively.

cd = 4 cm, so cd does not coincide.

So, point C and point D are a little between AB, and a little stupidity is on the BA extension. (otherwise coincides).

Let the interval between ab be c

ab=db-da=(9 5)da-da=(4 5)da, so da=(5 kai pants4)ab

ab=ac+bc=ac+(9 5)ac=(14 5)ac, so ac=(5 14)ab

cd=da+ac=(5/4)ab+(5/14)ab=(45/28)ab=4

ab=112/45

Let's say ad=ac=5

db=cb=9, which can be seen as an irregular and scattered, connecting ab, cd intersecting e, cd=4, and triangle adc and dcb are both isosceles triangles. then E is the midpoint of CD, and AB is perpendicular to DC. Use the Pythagorean theorem to find AE and Yuji EB respectively, and then add them together.

Calm and cautious.

The order of the four points of ABCD should be BCA D, because AD 5 9DB can be set to AD Liangkai 5T, AB=4T, and AC 5 Jing is changed to 9CB, BC=9S, CA=5S

Therefore, by cd=ca+ad=5t+5s=4(1), by ad-ac=5 9db-5 9cb=5 9(db-cb)=5 9cd=5 9*4=20 9

i.e. 5t-5s = 20 9 (2).

From equation (1) and (2), we can solve t=2 9+2 5s=2 5-2 9, so ab=bc+ca=9s+5s=14s=112 45

The order of the four dots on the Naoka line is the number of models CADB and set AD=5X

then db=9x

So ac=cd-ad=4-5x

cb=9/5ac=9/5(4-5x)

cb=cd+db=9x+4

9/5(4-5x)=9x+4

x=8 Suspicion of 45

ab=5x+9x=14x=14*8/45=72/45cm