What are the similarities and differences between to, for, and of? I am a 9th grader in junior high

to, as a preposition, is equivalent to carrying a rough "give", to is often followed by the town doThe original form of the verb, it is time to do (to do something) for as a preposition, often followed by doing, can also be a person with a missing limb (equivalent to "right... or for).

In high school, learning English is mainly for the college entrance examination.

Proficiency in the vocabulary manual of the college entrance examination is a must, method: insist on accumulating every day and take a certain amount of time.

The study of high school grammar is necessary, follow the explanation of the classroom teacher, and slowly understand it during the usual and holidays, and it is best to memorize some of it.

In the third year of high school, it is very important to do mock questions. While actually fighting, we also explored the idea of the problem.

Work hard, be willing to endure hardships, and the method is right = learn well.

[Listen more, you must listen more], this is the experience of many top English students. You can't be lazy with more backs.

If you have a good foundation in junior high school, it should not be difficult to learn high school English. If the foundation is not good, then it is best to make up for it, when your foundation is good, your next study will be easy, then you can learn well by reading more books and practicing more.

The most important thing is that the words are all floating clouds in high school English and everything.

Study hard, read more, write more, memorize more, and do more questions!!

Vocabulary, reading, listening, especially listening must be practiced well, it is very important to get to university!

Proof: Such a question.

It is a difficult problem, because the reason why it is called a theorem is that without it, it cannot be proved. In principle, theorems are not crossed, of course, there are some problems that will cross. It's all an isolated phenomenon.

If there is a crossover phenomenon in all theorems, it means that there is a duplication of theorems. One of these theorems is to be canceled. Therefore, the questions with restrictions are all difficult questions; Doing such questions does not help much to improve the academic level of mathematics.

The method of mathematics is the process of simplifying complex problems, rather than complicating simple problems. If you do too much of this kind of question, it will affect the way you think about it. For all college entrance examination answers, as long as you do the questions more easily, it means that your level is higher.

The more complex the question, the less clear your thinking is; Illustrate the less capable you are of the knowledge.

See the figure below, as of ab in f, isosceles rt aof and rt bof; as fh bo in h; Coupling EF, intersecting ob to g; This is the most straightforward and easiest way, however, there is no way to prove that the OGFH is square and the condition is missing. Therefore, it is proved with analytic geometry. Let ao=bo=4;According to the title:

1=∠2=45d/2=，∠adb=∠1+∠aob=90d+；Equation for straight line be:

y=tan∠adbx+4=;

The linear oe equation is: y=tan(-22,5d)=-[1 ( 2+1)]x=-( 2-1)x....2);

2)-(1), obtain: (-2+1)x+(1+ 2)x-4=0, x=2; y=-2(√2-1);point e, coordinates (2, 2-2, 2);

The linear equation of AE is: (y-0) (x-4)=(2-2 2-0) (2-4)=(2-1), and we get y=(2-1)x-4(2-1).3);

Because: -(2+1)=-(2-1) ( 2-1)=-1 ( 2-1): So, comparing the slopes of Eq. (1) and Eq. (3), the straight line ae be. The original proposition is proven. Certification.

The four-point contour is essentially the evolution of triangle similarity, so this problem can be proved by triangle similarity, as shown in Figure below

Counter-evidence can be used.

Assuming that AE is not perpendicular to BE, then the perpendicular line of A is the BE and the perpendicular foot is F (F does not coincide with E, F may be on DE or on the extension of DE). Extend the AF, and the extension line of the BO is at G.

Easy to prove that AFO=45 degrees.

Since f and e do not coincide, AFO and AEO are the relationship between the inner and outer angles of the triangle, and the two cannot be equal, contradictory!

So the original proposition holds.

Share a solution. Let = obe=, ob=a [analytically, if eoa= = eao can be proved, ea be can be proved]. Do EF OA in F.

By the question set the condition, there is eoa= efo= . Again, tan =. ∴od=(√2-1)a。

Instead, tan eoa=ef (od+df)=tan and tan eao=df ef=tan. Coupling, solution to get df = ( 2-1)od 2. of=od+df=a2, i.e., EF is the perpendicular bisector of OA.

eoa=α=∠eao。

EA BE was founded.

FYI.

Make ep=eo p on the bo extension line.

1= eod=

So eop= ep=eo

So epo= so peo=45°

So AE is perpendicular to AP

Or use the high school method to build the system and set the coordinates and set the equation of the straight line.