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Math topics (similar shapes)
1. Because CD∑AB, CD: AB = DE: EA;

Then there is DE:EA=3:6= 1:2, so DE: (DE+EA) =1:(1+2) =1:3, which means DE: DA =1:.

Because EF∨AB, DF:DB=EF:AB=DE:DA, that is EF:AB= 1:3.

Therefore, EF=2.

Agree 0| Comment

20 12-9-22 20: 27 Bob King 1990 | Know from the mobile phone | Level 3 1. The triangle ABD is similar to the triangle EFD, then the triangle CDB with EF/6=(4-BF)/4 is similar to EFB, and then EF/3=BF/4 is simultaneous. Two equations EF/AB=(BD-BF)/BD and EF/CD=BF/BD are obtained by similarity, and these two equations are added to obtain EF/AB+EF/CD= 1, that is, 1/ab+ 1/CD = 65438+

20 12-9-22 20:2 1 messy living sink | secondary (2)EF/CD=BF/BD EF/AB=DF/BD.

BF/4*3=(4-BF)/4*6

BF=8/3

Then EF=2.

1/AB = 1/6 1/CD = 1/3 1/EF = 1/2

So1/ab+1/cd =1/ef.

The first question is not parallel, so you can't agree with 0| comment.

20 12-9-22 20: 17 nearly clear | 7-level proof ∵EF∨AB

∴ DF/BD=EF/AB ①

∫EF∨CD

∴ BF/BD=EF/CD ②

①+②: 1= EF/AB+EF/CD。

1/EF = 1/a b+ 1/CD

Two problems to prove

AB = 6,CD=3

Substituting into the above formula,1/ef =1/6+1/3.

EF=2 Agree 0| Comment

20 12-9-22 20: 16 Zhao Yang 1990 130 | level 1 (2) EF/AB = DF/BD, ef/CD = BF/BDSOEF/AB+ef/CD.

20 12-9-22 20: 15 Di Hua Lian Demei | Secondary EF=2 Agree 0| Comment.

20 12-9-22 20: 13 enthusiastic user AB||EF So DEF is similar to DAB.

EF/AB=DF/DB

CD||EF So BFE is similar to BDC.

BF/BD=EF/DC

So EF/6=DF/4

BF/4=EF/3

BF+DF=BD=4

You can find EF=2.

2. by the same token, ef/ab = df/db; BF/BD = BE/CD;

The two formulas EF/AB+EF/CD=(DF+BF)/BD are added.

That is EF/AB+EF/CD= 1.

That is,1/ab+1/cd =1/ef agrees with 0| Comments.

20 12-9-22 20: 10 XM paladin | 13 level 1. Because CD∑AB, CD: AB = DE: EA;

Then there is DE:EA=3:6= 1:2, so DE: (DE+EA) =1:(1+2) =1:3, which means DE: DA =1:.

Because EF∨AB, DF:DB=EF:AB=DE:DA, that is EF:AB= 1:3.

Therefore, EF=2.

2.

Because EF∨AB, ef: ab = df: db; ( 1)

Because EF∑CD, ef: CD = BF: BD; (2)

( 1)+(2),get:ef:a b+ ef:CD = df:d b+ BF:BD =(df+BF):BD = 1。

So1/ab+1/cd =1/ef agrees with 0| comment.

20 12-9-22 20: 09 Liuping Haiyan | Grade 5 (1) ef/ab = ef/6 = df/4, EF/CD = EF/3 = BF/4 = (4-DF)/4SODF = 4/3.

(2) EF/AB = DF/BD, EF/CD = BF/BD SOEF/AB+EF/CD = (DF+BF)/BD =1Agreed.

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