Ethereum AE stolen

1. Example 1 is known: as shown in Figure 1-1, in the quadrilateral ABCD, BC > Ba, ad = CD, BD bisects ∠ ABC,

verification: ∠ a + C = 180 °

analysis: because the horizontal angle is equal to 180 °， Therefore, we should consider transforming two angles which are not together into equal angles by congruence, and there is no congruent triangle in the graph. Therefore, the key to solve the problem is to construct isosceles triangle, which can be realized by "truncation and complement method". "

prove: intercept be = AB on BC, connect De, and then take the midpoint m of EC, Connecting DM

∵ AB = be

and ∵ BD bisecting ∠ ABC a d

∵ abd = ∠ EBD

in △ abd and △ EBD,

AB = be

∠ abd = ∠ EBD

BD = BD B e M C

{abd ≌ △ EBD (SAS) is shown in Figure 1-1

} ad = ed ∠ a = ∠ bed,

∵AD = DC ， ∴ED = DC ∴∠ C = ∠DEC

∴∠A + ∠C = ∠BED +∠DEC = 180 °< As shown in Figure 2-2, AE / / BC, ad and BD share & eth equally; EAB、Ð CBA, EC, point D

verification: ab = AE + BC

analysis 1: to prove that ab = AE + BC, observe that AD and BD are bisectors, so the Daed can be folded along a, so it is necessary to add an auxiliary line to intercept BF = BC on AB, and only need to dece AF = AE, then the problem can be solved, so how to dece AF = AE becomes the key to solve the problem. Because the sum of internal angles of Daed, DADB and DBD is 180 °， And & eth; EDC=180 °， Because AE / / be, therefore & eth; E+Ð C=180 ° Thus & eth; EAB+Ð CBA=180 °， Ad and BD are bisectors of angles, and we can dece & eth; 1+Ð 4=90 °， Thus, we can dece & eth; ADB=90 °， Thus & eth; 6+Ð 8=90 ° If we can dece & eth; 7=Ð 8, then we only need to dece Daed ≌ dafd, thus we can dece AE = AF, BC = BF, & eth; 1=Ð 2. BD is a common edge, so we can dece dbfd ≌ DBCD, then & eth; 5=Ð 6, e to & eth; 5+Ð 7=90 ° So & eth; 6+Ð 7=90 °， Due to & eth; 6+Ð 8=90 °， Thus & eth; 7=Ð Therefore, ad can be the common edge, & eth; 3=Ð 4 dece Daed ≌ dafd, thus the idea is smooth, dece AE = AF, and dece AB = AE + BC by equivalent substitution

proof 1: intercept BF = BC on AB and link DF

BD is & eth; Bisector of ABC & eth; 1=Ð 2

in dbdf and DBDC,

(common side)

{dbdf ≌ DBDC (SAS) is shown in Figure 2-2

} & eth; 5=Ð 6 (the corresponding angles of congruent triangles are equal)

! & eth; 3+Ð 8+Ð E=Ð 4+Ð 1+Ð 5+Ð 7=Ð 2+Ð 6+Ð C=180 ° The sum theorem of internal angles of triangle)

{eth; E+Ð EAB+Ð ABC+Ð C+Ð EDC=540 °

and ﹣ AE / / BC ﹣ eth; E+Ð C=180 ° Two lines are parallel and complementary)

and ∵ & eth; EDC=180 ° ∴Ð 1+Ð 2+Ð 3+ Ð 4=180 °

ad is & eth; The bisector of EAB & eth; 3=Ð 4

∴Ð 1+Ð 4=90 ° ∴Ð 5+Ð 7=90 ° The sum theorem of internal angles of triangle)

{eth; 6+Ð 8=90 ° ∵Ð 5=Ð 6 ∴ Ð 7=Ð 8

in Daed and dafd,
{Daed ≌ dafd (ASA)

} AE = AF (the corresponding sides of congruent triangles are equal)

≌ AF + FB = AB

} AE = FB = AE + BC = AB

that is, ab = AE + BC


analysis 2: extend the extension line of BC intersecting ad to F. To prove AB = AE + BC, we only need to prove BF = AB and dece CF = AE. In order to prove CF = AE, we only need to dece two triangles Daed ≌ dfcd with CF and AE e to & eth; 5 = 6, AE / / BC, so & eth; 3=Ð F. If we want to dece ad = FD, it becomes the key to solve the problem °，& ETH; E+Ð BCE=180 °， Therefore, it can be known that & eth; EAB+Ð CBA=180 °， Ad and BD are & eth; EAB、Ð CBA's bisector, thus can be introced & eth; 1+Ð 4=90 °， Therefore & eth; ADB=90 °， Then & eth; EDB=90 °， At this point, they can dece dabd ≌ dfbd according to ASA by observing the figure, and thus dece ad = FD, forming the idea

proof 2: as shown in Fig. 2-3, the sum of the internal angles of the three triangles in Daed, DADB and DBDC is 540 ° The sum theorem of internal angles of triangles)

and ∵ & eth; EDC=180 ° Horizontal angle definition & eth; E+Ð C+Ð EAB+Ð ABC=180 °

AE / / BC  (two lines are parallel and complementary)

■ & eth; 3+Ð 4+Ð 1+Ð 2=180 °

AD and BD are & eth respectively; EAB、Ð The bisector of ABC

& eth; 3=Ð 4，Ð 1=Ð 2 (angular bisector definition)

﹣ eth; 1+Ð 4=90 ° ∴Ð ADB=90 ° The sum theorem of internal angles of triangle)

{eth; BDF=90 ° In DADB and dbdf,

≌ DADB ≌ dbdf (ASA)

≌ ad = FD, ab = FB, & eth; 4=Ð F (opposite sides of congruent triangle, corresponding angles are equal) as shown in Fig. 2-3

in Daed and dfcd, it is known that ad is the angular bisector of △ ABC, AB > AC, as shown in Fig. 3-1,

verification: AB-AC > bd-dc

analysis: in order to prove AB-AC > bd-dc, we need to transform the difference between AB and AC, the difference between BD and DC or their equal quantity into the edge of the same triangle, and then use the relationship between the three sides of the triangle to prove it

proof: Method 1: intercept AE = AC on AB and connect ed. A

∫ ad bisection ∠ BAC ∫ bad = ∠ DAC

in △ ade and △ ADC, E

AE = AC

ead = ∠ DAC B D C

ad = ad as shown in Figure 3-1

{△ ade ≌ △ ADC (SAS)} de = D C

in △ abd, Be > bd-de (the difference between the two sides of the triangle is less than the third side)

that is, ab-ae > bd-dc

Ψ AB-AC > bd-dc (equivalent substitution)

method 2: short complement method

extend AC to point E, so that AE = ab, Connecting de a

∫ ad bisection ∠ BAC ∫ bad = ∠ DAC

in △ bad and △ ead,

AB = AEC

∠ bad = ∠ DAC B D E

ad = ad

{ade ≌ delta ADC (SAS)} D B = D E, as shown in figure 3-2

in △ abd, EC > de-dc (the difference between the two sides of the triangle is less than the third side)

that is ae-ac > de-dc  AB-AC > bd-dc

example 4 shows that, as shown in Figure 4-1, in △ ABC, ∠ C = 2 ∠ B, ∠ 1 = 2.

verification: ab = AC + CD.

analysis: from the conclusion analysis, "truncation" or "complement" can realize the transformation of the problem, that is, extending AC to e makes CE = CD, It is proved that: Method 1 (short complement method)

prolongs AC to e, so that DC = CE, then ∠ CDE = ∠ CED, as shown in Figure 4-2

Figure 4-2

achb = 2 ∠ e,

∫ ACB = 2 ∠ B, ∫ B = e,


in △ abd and △ AED,

{abd ≌ △ AED (AAS)
, ≌, And the AE = AC + CE = AC + DC, < br /

< br /

< br /

< br / < br / < br / < br /
< br /
< br / < br / < br / < method 2 (truncation method) < br / < br /
in the Figure 4-3, as shown in Figure 4-3 < br / < br > < br / < br /
in △ AFD and △ ACD, < br /
< br /
< br / < br / < br > < br /
< br / < br / < br > < br / < br / < br > < br / < br / < br > < br / < br > < br / < br / < br > < br / < br / < br / < in △ AFD 87808780 8780 87808780b,

∴FD=FB. ∵AB=AF+FB=AC+FD，

∴AB=AC+CD.

2. At first, people were more interested in his design concept, but there was no shortage of fresh objects in the world. But once people were cut leeks, they would slowly think about this thing. He has been there for so many years, and it should graally return to zero, Even Ma Yun's technology application can't catch up, and it's still popular in the world. It's a joke. The state's attitude towards him is not to fight or support him.

3. I happened to see an activity in my circle of friends, BHT's new application for bitcoin. Now the market can continue to do activities, but I just want to continue to watch the market, have an in-depth understanding, and find something very interesting. The first time I saw BHT new, I was very confused about what it was. Internet search for BHT new didn't give much detailed explanation. Instead, it was stock new. I found that this BHT new came from stock new, which is not very different from stock new operation.

4. Currently, UOS is preferred. Because eth, EOS, AE have all gone up! UOS is the basic platform of blockchain which combines the advantages of eth + EOS.

5. The banyan essence incident in Guantou Central Primary School,
is false
for relevant information,
you can ask the school office and the academic affairs office
if you like,
it's sunny.

6. The possibility that swindler is bigger
recently, I posted a lot of his posters! And they are all trumpets of level 1
My Network account has been stolen, and there is his poster in the speech record! But I had to delete it myself
if you need and only have to register an account on his website, it is recommended that you do not fill in the account number, which is the same as other accounts! Do not rule out that they use the background user's data to hit the database
up to the user: Silk Entertainment's personal experience!

7. In order to protect your security, Unicom SIM card is protected by pin code and PUK code. When the pin code is wrongly input for three times, the SIM card will be locked automatically, and the mobile phone will prompt "SIM card is locked"
after the SIM card is locked, you need to input PUK to unlock it. You can log in to the mobile phone business hall and click "my Unicom" > "security center" > "my PUK code" to query the PUK code

warm tips:
1. The initial password of pin code is 1234. If the PUK code is input incorrectly for 10 times in a row, the SIM card will be scrapped and can only be used after the card is replaced
2. The specific functions and operation path of mobile phone business hall are subject to the actual page.

8. There's no such thing, right? With what? Pay attention to the password when surfing the Internet, don't steal it, it may be that aspect

9.

Press the shortcut key y to switch to the anchor tool,

and drag the anchor to the center of rotation you want