Auxiliary line, how to add it? Master theorems and concepts.
We should study hard and find out the rules by experience.
triangle
There is an angular bisector in the picture, which can be perpendicular to both sides.
You can also look at the picture in half, and there will be a relationship after symmetry.
Angle bisector parallel lines, isosceles triangles add up.
Angle bisector plus vertical line, try three lines.
Perpendicular bisector is a line segment that usually connects the two ends of a straight line.
[Example 1]
As shown in figure 1, d is the midpoint of AC on the side of ⊿ABC, and BC is extended to point E, so that the extension lines of CE = BC and ED intersect with AB at point F, and ED∶EF is found.
Analysis:
Idea 1: For the parallel line that passes through C and makes AB intersect G, FD=DG can be obtained by taking D as the midpoint of AC, and FG=GE can be obtained by taking CE=BC, thus ED∶EF=3∶4 can be obtained.
Idea 2: Parallel lines intersecting D and BE and AB intersecting I are similar to Method 1, with ID: BC =1:2 and ID: Be =1:4, so that ED: ef = 3: 4.
Idea 3: When the parallel line intersecting D and AB intersects H, BH=HC= 1/4BE and ED∶EF=3∶4 are easily obtained.
Note: The addition of three parallel lines in the three ways of thinking in this question makes full use of the condition that "D is the midpoint of ⊿abc AC side", which makes an originally weak condition rich in connotation under the action of parallel lines, which not only appears the midpoint of the other side, but also makes use of the midline theorem of triangles, making it more handy to use.
Sometimes it is necessary to add some graphs to construct graphs to make up for the deficiency of the topic, so that the conditions of the topic can be fully displayed, thus creating conditions for the application of the theorem, or transforming a conclusion that cannot be directly proved into another conclusion that is equivalent to it, which is convenient for thinking and proving.
[Example 2]
It is known that O is a point in the square ABCD, ∠ OBC = ∠ OCB = 15. Prove that ⊿AOB is an equilateral triangle.
Analysis:
(as shown in Figure 2) Construct a triangle OMC. Let DH⊥OC be H, then ∠ 2 = 15 ∠ DCM = 15, then ⊿DMC≌⊿BOC and ∠ MCO = 60.
∴∠ DMO = 360-60-150 =150 ∴∠∠1= ∠ Doc =15, so there is ∠.
Description: This problem is to construct an equilateral triangle similar to the conclusion to be proved with auxiliary lines, and then solve the problem with the help of the constructed graphics.
Gather scattered geometric elements together.
In some geometric problems, conditions and conclusions are scattered. By adding appropriate auxiliary lines, the scattered and "distant" elements in the diagram are gathered on the related diagram, so that they are relatively concentrated, which is convenient for comparison and relationship establishment, so as to find out the solution to the problem.
[Example 3]
As shown in Figure 8, the bisector of ∠B=2∠C and ∠A in △ ABC is AD. Is the sum of AB and BD equal to AC?
Idea 1: As shown in Figure 9, AE=AB is intercepted on the long line segment AC, and BD=DE is deduced from △ Abd △ AED, so it is only necessary to prove EC=DE.
Idea 2: As shown in figure 10, extend the short line AB to point E to make AE=AC, then you only need to prove BE=BD, and you can prove ∠E=∠BDE from △ AED △ ACD and ∠B=2∠C, so there is BE=BD.
Idea 3: As shown in figure 10, extend AB to E, make BE=BD, connect ED, from ∠ABD=2∠C, ∠ABD=2∠E, prove △ AED △ ACD, and AE=AC, that is, AC.
Explanation: In this example, line segments AB and BD that are not on a straight line are combined into a straight line with auxiliary lines, so that AB+BD or AC-AB can be easily obtained, and then the problem can be solved.
The method of adding auxiliary lines in plane geometry is flexible and changeable, which requires us to master the basic concepts and theorems in mathematics, often classify and summarize them in practical exploration, carefully analyze the conditions given by the questions, and find some implicit and regular information from them.
source
It needs to be proved that the line segment is double-half, and extension and shortening can be tested.
The two midpoints of a triangle are connected to form a midline.
A triangle has a midline and the midline extends.
quadrilateral
A parallelogram appears and the center of symmetry bisects the point.
Make a high line in the trapezoid and try to translate a waist.
It is common to move diagonal lines in parallel and form triangles.
The card is similar, and it is a habit to add lines parallel to the line segment.
It is very important to find the line segment in equal product proportional conversion.
Direct proof is more difficult, and equivalent substitution is less troublesome.
Make a high line above the hypotenuse, which is larger than the middle term.