Geometry is the most difficult and flexible part in junior high school mathematics, because almost all the contents of Euclid's plane geometry have been learned in junior high school. The content and coverage are enough to make it confusing and unfathomable. Making auxiliary lines is the key to solving many geometric problems, and most beginners are confused about it. Over time, they felt afraid and bored. In fact, even experienced teachers. When you encounter a new problem, you may not be able to solve it immediately. So novices don't have to have too much psychological pressure on it. Remember, as long as you pay, you will get something back. Now there is a saying that geometry problems need a kind of feeling, which is called geometry feeling. Why do auxiliary lines do this instead of that? I don't know, because it depends on the feeling. I don't object to the idea of geometric sense, but it's a bit much to say that there is no reason for it. Every questioner has his thoughts and intentions in every question. How can you say there is no reason? Besides, every auxiliary line is not made out of thin air. So I suggest that every beginner should seriously think about it and guess the questioner's intention when doing geometry problems. But he was not born, but earned it with sweat. I believe that those who can solve geometry problems by feeling must have paid a lot, but I think he can do better by thinking more about each problem.
It seems that we have gone too far. Here are some techniques for making auxiliary lines. But before that, I have to remind you that any skill and method can't be separated from knowledge. Without strong basic knowledge as the backing, everything will be an armchair strategist. So it can be said that learning every theorem in the book is the key to everything. Did you do it? If not, please find a new notebook immediately and try to prove two theorems in the book from beginning to end every day. Instead of reading the proof in the book, it is better to find a way to write the process in detail, and each theorem is accompanied by two related exercises.
Question (you can find reference materials), stick to it, and finally I believe you can find the so-called geometric sense.
Let's look at a few examples first.
1. As shown in the figure, the quadrilateral ABCD is rectangular, and BF⊥DE is verified in F: AF⊥FC.
Analysis: The topic is very simple, and seeking truth is illusory. Observe, observe, observe again,
To prove AF⊥FC is actually to prove that the triangle AFC is a right triangle, and then consider
The diagonal lines of a rectangle are equal and equal, and the four corners are right angles. So you should be able to figure out how to make auxiliary lines.
Let's go If you don't have a clue, look again. The fact that so many line segments are equal (in a rectangle) seems to imply that the median line on the hypotenuse of a right triangle is equal to half of the hypotenuse. Since then, the problem has become clear.
Proof: Connect AC, BD and FO.
Because BO=OD and FO are the center lines on the hypotenuse of right triangle BFD, FO=BO=OD= 1/2BD.
AC=BD, AO=OC, so FO= 1/2AC, so the center line FO of AC side in triangle AFC is equal to half of AC, so the angle AFC=90 degrees. I.e. AF vertical FC
After the proof is completed, in retrospect, how simple and clear the process is depends on the auxiliary credit investigation. It seems that this problem is not easy to solve in other ways. Therefore, it is often the key to solve the problem to do a good job and correct the auxiliary line. It can even be said that the assistant was done first and the topic was half done. I hope the students will learn more!
As shown in fig. 2
In a square ABCD, AC and BD BD intersect at point O, FA bisects angle BAC, DB at point E, and BC at point F. It is proved that OE= 1/2FC.
Analysis: This is an old problem, but it is a problem of cultivating ability. Making auxiliary lines has also become the key. How? Think about it. If you can find a line segment, it is half of FC and half of OE. Obviously, you need to find a line segment equal to half of FC. Just cross O to make OH//FC intersect AF in H, and OH is obviously the center line of triangular AFC, so there is OF= 1/2FC, which is the line segment you are looking for. Just prove it = below. The practice of auxiliary line in this question is simpler than the previous one. You should be able to see it at a glance, unless you don't know the mean value theorem, so you need knowledge to have skills. That is to say, it is often emphasized that students should lay a good foundation.
It is proved that o is OH//FC and AF is h,
It is easy to know that OH is the center line of the AFC Triangle, ∴OH= 1/2FC.
∠Ohe =∠EFB =∠ACF+∠FAC = 45+∠Fab =∠Abd+∠Fab =∠BEF =∠OEH。
∴OH=OE
∴OH= 1/2FC.
The proof process of this problem is linked one by one, and the whole proof process is like helping each other in the same boat. There are other ways to do the auxiliary line of this problem. Can you think of it?
3 as shown in fig. 3, in triangle ABC(ab >；; Take a point D on the side of AB and a point E on AC, so that
AD=AE, the extension line of straight line DE intersects BC at point P. Verification: BP:CP=BD:CE,
Analysis: Considering this problem, prove that the line segments are proportional, and think of parallel lines with similar shapes. Obviously, this problem is a parallel line. The intersection point c is CM//AB and PD is m,
have
BP/CP=BD/CM，AE/CE=AD/CM，
And AD=AE, CM=CE,
So BP/CP=BD/CE,
There are many ways to use this question as an auxiliary line, and each method is a method. If point B is BM//DP, and the AC extension line is in M, or point B is BM//CA, and the PD extension line is in M, students may wish to give it a try, but no matter how the auxiliary line is made,
Be sure to catch BP/CP, so as to be effective.
Through the above questions, I simply told the students some specific methods of making auxiliary lines, but due to the limitation of space, I can't tell them all.
For example, the practice of auxiliary lines is flexible, requiring students to analyze specific problems, and different topics have different ideas. I can only give students some basic things, that is, the idea of making auxiliary lines As long as they are willing to use their brains, there are ways to solve the problem.
One last example:
3 as shown in fig. 3, in triangle ABC(ab >；; AC) Take a point D on the side of AB and a point E on AC, so that AD=AE, the extension line of straight line de and BC intersect at point P. Verify: BP:CP=BD:CE,
Analysis: Considering this problem, prove that the line segments are proportional, and think of parallel lines with similar shapes. Obviously, this problem is a parallel line. The intersection point c is CM//AB and PD is m,
have
BP/CP=BD/CM，AE/CE=AD/CM，
And AD=AE, CM=CE,
So BP/CP=BD/CE,
There are many ways to use this question as an auxiliary line, and each method is a method. If point B is BM//DP, AC extension line is in M, or point B is BM//CA, and PD extension line is in M, students may wish to give it a try, but no matter how the auxiliary line is made, it is necessary to grasp BP/CP to work.