List of high-quality authors
What I would like to share with you is: Junior High School Mathematics|Summarize the 4 common methods of constructing auxiliary lines, collect and reserve! I hope it will be helpful to you by providing ideas in your daily study and answering questions in exams.
1. About the auxiliary line of the angular bisector.
When the condition of the problem is bisectoral, it is necessary to think of constructing auxiliary lines according to the nature of the angular bisector.
The angular bisector has two properties: the angular bisector has symmetry; The distance from the point on the bisector of the angle to both sides of the angle is equal.
Commonly used auxiliary line methods for angular bisector lines:
(1) Interception constructs congruence
As shown in the figure, OC is the angular bisector of AOB, D is a point on OC, F is a point on OB, if we take a point E on OA, so that OE=OF, and connect DE, then there is OED OFD, thus creating conditions for us to prove that the line segments and angles are equal.
(2) The points on the angular equinox line are made perpendicular lines on both sides of the angle
The problem is proved by the property that the distance from the point to both sides on the bisector of the angle is equal. As shown in the figure, OA and OB on both sides of the angular angle of one point D on the bisector OC through AOB are perpendicular, and the perpendicular foot is E and F, connecting DE and DF.
Then there are: de=df, oed ofd.
(3) The perpendicular line of the angular bisector constructs an isosceles triangle
As shown in the figure, from a point E on one side of the angle ob as the perpendicular line EF of the angle bisector OC, so that it intersects with the other side of the angle OA, then an isosceles triangle (OEF) is intercepted, and the vertical foot is the midpoint D on the bottom edge, and the angle bisector becomes the midline and height on the bottom edge, so as to take advantage of the property of the median line and the three-line integration of the isosceles triangle.
If there is a line segment perpendicular to the bisector of the angle, extend the line segment to intersect with the other side of the angle, thus giving an isosceles triangle, which can be summarized as: "Extend the pendulum, isosceles return".
(4) Make parallel lines to construct an isosceles triangle
The isosceles triangle constructed as a parallel line is divided into the following two cases:
As shown in the figure on the left below, a point E on the bisector oc is used as a parallel line de on one side of the angle OA, thus constructing an isosceles triangle ODE.
As shown in the figure on the right below, the parallel line DH of the angular bisector OC is made by the point D on one side of the angle ob and the opposite extension line of the other side AO intersects at the point H, thus constructing an isosceles triangle ODH.
2. Auxiliary lines thought of by line segments and differences
1) When it is found that one line segment is equal to the sum of the other two line segments, the general method is the truncation method
Truncation: Truncating one segment of a long line is equal to one of the other two, and then proving that the remaining part is equal to the other;
Fill in the short: Extend a segment so that the extension is equal to another, and then prove that the new segment is equal to the long segment.
Truncation method as auxiliary line:
Examples:In ABC, AD bisects BAC, ACB 2 B, and verifies ab ac CD.
2) For proving the inequality of the sum of the line segments, it is usually related to the fact that the sum of the two line segments in the triangle is greater than the third side and the difference is less than the third side, so it can be proved in a triangle.
When using the trilateral relationship of the triangle to prove the unequal relationship of the line segments, if it cannot be proved directly, two points or a certain side of the court length can be connected to form a triangle, so that the line segment of the conclusion is in one or several triangles, and then the unequal relationship of the three sides of the triangle is used to prove.
Examples:It is known as Figure 1-1: D and E are two points in ABC, and the verification is ab ac bd de CE
3) When the outer angle of the triangle is greater than any inner angle that is not adjacent to it, if it cannot be directly proven, two points can be connected or a certain side can be extended to construct a triangle, so that the big angle of the triangle is in the position of the outer angle of a triangle, and the small angle is in the position of the inner angle of the triangle, and then the outer angle theorem is used
For example:As shown in Figure 2-1, if D is known to be any point in ABC, verify BDC BAC.
Analysis: Because BDC and BAC are not in the same triangle and have no direct connection, auxiliary lines can be added to construct a new triangle, so that BDC is in the position of the outer corner and BAC is in the position of the inner corner.
Evidence 1:Extend BD at point E, where BDC is the outer angle of EDC, BDC Dec, and the same is true for Dec Bac, BDC Bac
Evidence 2:Connect AD and extend the BC to F
BDF is the outer angle of ABD.
BDF Bad, the same goes for CDF CAD
bdf＋∠cdf＞∠bad＋∠cad
Namely: BDC BAC.
Note: When using the outer angle theorem of a triangle to prove the inequality relationship, the large angle is usually placed at the position of the outer corner of a triangle, and the small angle is placed at the position of the inner corner of the triangle, and then the inequality property is proved.
3. The auxiliary line that comes from the midpoint
In a triangle, if a point is known to be the midpoint on one side of the triangle, then the first thing to think of is that the midline of the triangle doubles and extends the midline and its related properties (the property of the bottom midline of an isosceles triangle), and then explores to find a solution to the problem.
(1) The middle line divides the original triangle into two small triangles of equal area
That is, as shown in Figure 1, AD is the midline of ΔABC, then SδABD=SδACD=1 2SδABC (because ΔABD and ΔACD are the same height as the bottom).
(2) Double-length midline
From the nature of the midline, it can be seen that a middle line divides the line segment where the midpoint is located, and a group of equilateral and opposite angles can be obtained through the double length of the midline, so that a group of congruent triangles can be obtained.
Examples:As shown in Figure 3, in isosceles ABC, AB=AC, BD is intercepted on AB, CE is intercepted on AC extension line, and CE=BD is connected to DE and BC is verified by F: DF=EF
4. Other auxiliary line practices
(1) Extend the triangle of the known edge
In some verification triangle problems, extending the intersection of two line segments (edges) to form a closed graph can find more equality relations and help solve the problem
Examples:As shown in Figure 4, in ABC, AC=BC, B=90°, BD is the bisector of ABC If the distance AD from point A to the straight line BD is A, find the length of BE
(2) Connect the diagonal lines of the quadrilateral and solve the problem of the quadrilateral into a triangle.
(3) Connect known points to construct congruent triangles
Examples:Known: Figure 10-1; AC and BD intersect at point O, and AB DC, AC BD, verify: A D.
(4) Take the midpoint of the line segment to construct a congruent triangle
For example, as shown in Figure 11-1, ab dc, a d verify: abc dcb.
Analysis: From ab dc, a d, think of taking the midpoint n of ad, connect nb, nc, and then have abn dcn by the sas axiom, so bn cn, abn dcn. Now we only need to prove NBC NCB, and then take the midpoint M of BC and connect MN, then there is NBM NCM by the SSS axiom, so NBC NCB. The problem is proven.
*:The content of the article is on the Internet, compiled and edited by the parents of junior high school students, you are sharing and exchanging learning, and the copyright belongs to the original author and the original source.