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As shown in the figure, given that point E is on the edge ab of △ ABC, point D is on the extension line of Ca, and point F is on the extension line of BC, what is the size relationship between ∠ ACF and ∠ D? Please give reasons
The reason is as follows: ∵ BAC is the outer angle of △ ade, ∵ BAC ＞ D, ∵ ACF is the outer angle of △ ABC, ∵ ACF ＞ BAC, so ∵ ACF ＞ D
As shown in the figure, in △ ABC, extend CA to e, extend BC to F, and D is a point on ab
It is proved that: ∵ ACF ﹥ cab, ﹥ cab ﹥ ade, ﹥ ACF ﹥ ade
Given that there are two points c and D on the line AB, and AC: CD: DB = 2:3:4, e and F are the midpoint of AC and DB respectively, EF = 2.4cm, the length of line AB is calculated
Answer by 3:00 this afternoon
Let AC = 2x, CD = 3x, DB = 4x
∵ E and F are the midpoint of AC and DB respectively
∴EF=½AC＋CD＋½DB=x＋3x＋2x=6x
∵EF=2.4cm
∴x=0.4
∴AB=AC＋CD＋DB=9x=3.6cm
Given that there are two points c, D, e and F on AB, which are the midpoint of AC and DB respectively, EF = 2.4cm and CD = 1.2cm, find the length of ab
(2.4-1.2)*2+1.2=3.6(Cm)
Points c and D are two points on line AB, AC: CB = 2:3, ad: DB = 4:1. If CD = 3cm, how many cm is ab
AC:3+DB=2:3
AC+3:DB=4:1
AC=3,DB=1.5,AB=3+3+1.5=7.5
As shown in the figure, point CD is two points on line AB, and AC: CB = 5:7, AD / DB = 5:11. If CD = 10 cm, find the length of ab
AC:CB=20:28(20+28=48),AD:DB=15:33(15+33=48).AD:DC:CB=15:5:28,AB=(10/5)*(15+5+28)=96cm
There are two points in the line AB: C, d.ac: CB = 2:5, ad: DB = 5:6, CD = 13?
∵ac:cb=2:5,ad:db=5:6
∴ac=2/7ab,cb=5/7ab；ad=5/11ab,bd=6/11ab
∵ad=ac+cd,cd=13
∴5/11ab=2/7ab+13
∵13=13/77ab
∴ab=77
Note: you are in grade two of junior high school, right? You haven't learned the linear equation of two variables. This is the standard way to solve the line segment problem
Given that there are two points c and D on the line AB, and AC: CB = 2:5, ad: DB = 5:6, CD = 13, find the length of the line ab
fast
∵（a-10)²+│b/2-4│=0
∴a-10=0 b/2-4=0 ∴a=10 b=8
∴AB=10 AC=BD=8
∴AD=BC=2
∵ m and N are the midpoint of AC and ad respectively
∴AM=1/2AC=4 AN=1/2AD=1
∴MN=AM-AN=3
C is the midpoint of line AB, D and E are the points on line AC and CB respectively, and ad = two-thirds AC, de = three fifths AB, ab = 24cm
C is the midpoint of AB, then AC = CB = 12cmad = 2 / 3 AC, then 12 times 2 / 3 = 8cm, that is, ad = 8cmad plus DC = 12, then DC = 12-8 = 4cm, de = 3 / 5 AB, that is, 24 times 3 / 5 = 14.5cm = De, known DC = 4, de = 14.5, then CE = 14.5-4 = 10.5cm, EB = 12-10.5 = 2.5cm
It is known that: C is the midpoint of line AB, D is the point on CB, e is the midpoint of DB, if CE = 4, ad = 2 / 3 AB, find the length of line ab
It is known that: C is the midpoint of AB, so: AC = 1 / 2Ab, and because: ad = 2 / 3AB and AC + CD = ad, so: CD = ad-ac = 2 / 3ab-1 / 2Ab = 1 / 6ab. In addition: BD = ab-ad = 1 / 3AB and E is the midpoint of BD, then de = 1 / 2bd = 1 / 6ab. To sum up: CE = CD + de = 1 / 6ab + 1 / 6ab = 1 / 3AB = 4cm, then 1 / 3AB = 4cm, ab = 12cm
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