As shown in the figure, in the quadrilateral ABCD, ∠ bad = ∠ ABC, ad = BC, AC intersects with point o at BD. let's say that the triangle OAB is an isosceles triangle

As shown in the figure, in the quadrilateral ABCD, ∠ bad = ∠ ABC, ad = BC, AC intersects with point o at BD. let's say that the triangle OAB is an isosceles triangle

In △ abd and △ ABC
AB=AB
∠BAD=∠ABC
AD=BC,
△ABD≌△ABC
∠BAC=∠ABD
The triangle OAB is an isosceles triangle

As shown in the figure, in rectangle ABCD, AB: BC = 1:2, point E is on edge ad, and 3aE = ed

∵ ABCD is rectangular, ∵ EAB = ∠ ABC = 90 °, BC = ad. ∵ AB: BC = 1:2, 3aE = ed, ∵ AE: ab = 1:2. ∵ ABC ∽ EAB

It is known that: as shown in the figure, in △ ABC, BP and CP divide ∠ ABC and ∠ ACB equally, de passes through point P, intersects AB at D, intersects AC at e, and de ‖ BC. Verification: de-db = EC

∵ BP bisects ∠ ABC, ∵ DBP = ∠ CBP. ∵ de ‖ BC, ∵ CBP = ∠ DPB. ∵ DPB = ∠ DBP. That is, DP = dB. Similarly, PE = CE. ∵ de = BD + CE, that is, de-db = EC

In △ ABC, ab = 14, point E is on AC, point D is on AB, if AE = 3, EC = 4, and AD / DB equals AE / EC, is it true to find ad length dB / AB = EC / AC

ad/DB=AE/EC=3/4
Ad = 3 / 4dB = 3 / 4 (14-ad), ad = 6
Yes, DB = 14-ad = 8, DB / AB = 8 / 14 = 4 / 7
EC/AC=4/(3+4)=4/7

As shown in the figure, point C is the trisection point of line AB, point D is on CB, CD: DB = 17:2, and cd-cb = 3cm, find the length of line ab

Let CD = 17x, so BD = 2x
Because C is the trisection of ab
So AC = BC / 2
So AC = 19x / 2
Because cd-ac = 3
So 17x-19x / 2 = 3
The solution is x = 2 / 5
So AB = 19x / 2) * 3 = 57 / 5

Point C is the trisection point of line AB, point D is on CB, CD: DB = 17:3, and CD AC = 7cm

Please discuss according to the position of C and D, as follows:
1、 When C is close to point a and D is close to point B:
Set the midpoint of BC as m, because C is the trisection of AB, so AC = cm = MB
Because cd-ac = 7, cd-cm = 7, that is MD = 7. Because CD: DB = 17:3, CD = 1 / 3AB + 7; DB = 1 / 3ab-7, CD: DB = (1 / 3AB + 7): (1 / 3ab-7) = 17:3
Results AB = 30
2、 When both C and D are close to point B:
Because cd-ac = 7, CD > AC. does not hold
3、 When C is close to point B and D is close to point A:
Because CD: DB = 17:3, so CD > dB. Is not true!

Point C is the trisection point of line AB, point D is on CB, CD: DB = 11:2
And cd-ac = 3cm, find the length of line ab

AC=1/3AB -----------(1)
Given that point D is on CB, CD + DB = CB = 2 / 3AB, and CD: DB = 11:2
So, CD = 22 / 39ab (2)
It is known that cd-ac = 3cm, combined with formula (1) and formula (2), 22 / 39ab-1 / 3AB = 3cm
The solution is ab = 13cm

As shown in the figure, if C is the midpoint of line AB, D is on line CB, Da = 6, DB = 4, then the length of CD is______ ．

Because Da = 6, DB = 4, so AB = 10, because C is the midpoint of AB, so AC = 5, because Da = 6, so CD = 1

The points c and D are on the line AB, and the lengths of the segments AC, CB, ad and DB satisfy
Points c and D are on the straight line ab. the lengths of the segments AC, CB, ad and DB satisfy AC: CB = 5:4, ad: DB = 2:1, and CD = 2cm. Find the length of the segment ab,

Let the length of AB be x cm in the first case: --- --- --- a C D BAC = 5 / (5 + 4) AB = 5 / 9 abad = 2 / (2 + 1) AB = 2 / 3 ABCD = ad - AC = (2 / 3-5 / 9) AB = 1 / 9 AB = 2Ab = 18 cm in the second case

C is the point on line AB, D is the midpoint of line AC, ab = 20cm, DB = 16cm, find the length of line CB

Let CB length YCM, DC = ad = AC / 2 = x be solved
X+y=16 ①
2X+y=20 ②
②—①
X=4
∴y=12
CB length 12cm