Even so, the topic can be regarded as making a square AEHI with H as the moving point and he as the side length, moving on GF. Because point H is moving, the corresponding point A will also move with it, but I can assure the landlord that when point H moves to point G, at this moment AG=2BG or AB=BG, because they satisfy the vertical relationship between them. When point h moves to point f, point a coincides with point B.
According to the above point of view, the side length of square ABCD is changing. This satisfies the premise that A 2+B 2 = C 2, that is, if I give you a quantitative value (assuming B), then the value of A will change with the change of C, and the range of AB must be between (0-b) due to the limitation of the movement of H point on GF.
If the landlord just thinks that A, B and C are side length comparisons, then this question is meaningless. The key point is that Liu Hui adopted the idea of extension, that is, square AEHI= square ABCD+ square BEFG.
After reaching the above knowledge, we will discuss the significance of generalization to Pythagorean theorem. Let's assume that the intersection of AG and IH is O.
Ideas:
First of all, the landlord has restricted me from using congruence, which is based on the proportional relationship between parallel lines. We rotate the triangle HEF by 90 degrees. Make points a, e and h on the same straight line. At this point, B, E and F are also on the same straight line, HF//AB, so we can get HF=AB=a (this is not congruence, it is the concept of parallel lines, and the line segments are proportional). You can also think of this as extending the triangle HEF to the right of the whole graph.
In fact, as long as AB=HF=a, then GH=b-a, this problem is almost solved. We assume that the intersection of AG and IH is O, then according to the concept that line segments between parallel lines are proportional, we can express line segments OG, GH and OH as A and B respectively.
At this time, the area of square AEHI is equivalent to the sum of the areas of triangle AIO, triangle Abe, triangle BOE and triangle EOH, and these triangles are right triangles, and each side length can be represented by algebra A and B, while the area of square AEHI itself is C 2, so there is no need to worry about identity. (Note that as long as you add up the areas of four right triangles, you don't need to make equations, because they are equal and will definitely be the concept of identity, that is, A-B=0).
Syndrome closure
The key is to see the essence from this question, in fact, that is, the H point moves on GF and asks you the relationship between the side length AB and HF of the square. Once we grasp the essence, it is easy to think clearly about this problem.
This is the same as many situations we encounter in our lives. Seeking a clear train of thought and seeing the essence of the problem is very helpful to improve your efficiency and stand out from the crowd.