In △ ABC, ab = AC, be = CF, EF intersect BC with D

In △ ABC, ab = AC, be = CF, EF intersect BC with D

Make FG / / AB cross BC extension line at G
Then ∠ g = ∠ B
From ab = AC, it can be known that: ∠ B = ∠ ACB
However, ACB = GCF
So, g = GCF,
So, CF = GF
And CF = be
So be = GF
∠G=∠B
∠BDE=∠GDF
Therefore, △ BDE ≌ △ GDF
DE=DF

As shown in the figure, D and F are the points on △ ABC edges AB and AC respectively, and ad: DB = CF: FA = 2:3, connecting the extension line of DF intersection BC edge to e, then EF: FD=______ ．

If FG ∥ AB is made by F and BC is made by G according to CF: FA = 2:3, then fgab = cfac = 25 ∥ FG = 25ab according to AD: DB = 2:3 ∥ bdab = 35 ∥ BD = 35ab ∥ fgbd = 23 ∥ efed = fgbd = 13 ∥ effd = 2

As shown in the figure △ ABC, BD ⊥ AC in D, CE ⊥ AB in E, ad = DH = 1, CD = 5, then the area of triangle ABC is
It's tonight

∵BD⊥AC,CE⊥AB
∴∠BDA＝∠BDC＝∠AEC＝90
∴∠ABD+∠A＝90,∠ACE+∠A＝90
∴∠ABD＝∠ACE
∵AD＝DH＝1
∴△ABD≌△HCD （AAS）
∴BD＝CD＝5
∵AC＝AD+CD＝1+5＝6
∴S△ABC＝AC×BD/2＝6×5/2＝15

In the triangle ABC, the bisector ad of ∠ ACB = 90 ° intersects BC at point D, CE ⊥ AB intersects ad at point F, AB at point E, DH ⊥ AB at point H
Verification: the quadrilateral cdhf is a diamond

According to the meaning of the title, CF ‖ DH
Because ad = ad ∠ CAD = ∠ had, RT △ ADC ≌ RT △ ADH CD = DH ∠ HDF = ∠ CDF
Therefore, ∠ HDF = ∠ CFD = ∠ CDF △ CFD is isosceles △ CF = CD
It can be seen that CF and DH are parallel and equal, and the adjacent side CD = DH

Given the line segment CD, extend CD to B so that DB = 1 / 2CD, extend CD to a so that AC = CD, if AB = 10 cm, then CD =?
Given the line segment CD, extend CD to B to make DB = 1 / 2CD, and then extend CD to a reversely to make AC = CD. If AB = 10 cm, then CD =?

Are you sure you didn't copy the wrong question?
Don't miss the reverse
Because AB = AC + CD + dB and AC = CD = 2dB
So AB = 2dB + 2dB + DB = 10cm
It can be 5dB = 10cm
So DB = 2 cm
So the final result is AC = CD = 2dB = 4cm

Given the line segment CD, extend CD to B so that DB = 1 / 2CB, and then extend DC to a so that Ca = BD. if AB = 15cm, then CD =?

By extending CD to B, DB = 1 / 2CB, DB = CD. By extending DC to a, CA = BD, CA = BD= CD.AB=AC +CD + DB = 15cm. Then CD = 5cm. Do your own drawing

As shown in the figure, DC ⊥ Ca, EA ⊥ Ca, CD = AB, CB = AE

As shown in the figure, ∵ DC ⊥ Ca, EA ⊥ Ca, ∵ C = ≌ a = 90 °, in △ BCD and △ EAB, CD = ab ≌ C = ≌ DCB = AE, ≌ BCD ≌ EAB (SAS)

Given the line segment CD, extend CD to B, so that DB = 1 / 3CD, extend DC to a, so that Ca = 1 / 2CB, if AB = 12, find the length of CD

cd=3db
cb=cd+db=4db
ca=1/2cb=2db
ab=ac+cd+db=2db+3db+db=6db=12
db=2
cd=3*2=6

AC vertical CB, DB vertical CB, ab = DC
Question 8 of 2

Known: two triangles are right angles
We can prove congruence with HL
Because the two right angles are equal, so you ask which two angles are equal
I'm sorry to send it by mobile phone

As shown in the figure, it is known that C is the midpoint of AB, D and E are the points on the line segments AC and CB respectively, and ad = 2-3ac, de = 3-5ab. If AB = 12, the length of the line segment CE is obtained

ce=de-dc
dc=ac-ad
So CE = de AC + ad
C is the midpoint of ab
ac=1/2*ab=1/2*12=6
ad=2\3ac,ad=2/3*6=4
de=3\5ab,de=3/5*12=7.2
So, CE = 7.2-6 + 4 = 5.2