Translation, rotation and symmetry are three transformations in plane geometry, and adding auxiliary lines by translation, rotation and symmetry is the basic idea and common method to solve the problem of geometric proof. While analyzing the characteristics of graphics, it is very important to guide students to master the appropriate method of adding auxiliary lines to improve their problem-solving ability.
2. 1 Add auxiliary lines by translation.
Speaking of trapezoid, the waist or diagonal of trapezoid is often translated into parallelogram and triangle.
Example 1. In trapezoidal ABCD, DC∨AB, ∠A and ∠B are complementary, and M and N are the midpoint of DC and AB respectively. Prove that MN=(AB-CD).
Analysis: translating DA into ME and CB into MF will form □AEMD□BFMC and □EMF. It is easy to prove that △EMF is a right triangle and MN is the midline on the hypotenuse EF, so MN=EF, EF=AB-CD. Of course, it can also be done by adding other auxiliary lines, but this is faster.
Example 2. In trapezoidal ABCD, AD∨EF∨BC, AD= 12, BC= 18, AE∶EB=2∶3, find the length of EF.
Analysis: Let point D be DG∨AB, and hand in EF to H and BC to G respectively, as long as the lengths of EH and HF are calculated respectively.
Solution: Point D is DG∨AB, EF is crossed by H and BC by G respectively.
∫AD∨EF∨BC，AD= 12，BC= 18，
∴ad=eh=bg= 12 ∴gc=bc-bg= 18- 12=6
AE∶EB= DH∶HG=2∶3 ∴DH∶DG=2∶5
∫DH∶DG = FH∶CG ∴fh∶6=2∶5
∴fh=2.4 ∴ef= 12+2.4= 14.4
2.2 Use Rotation to Add Guides
2.2. 1 refers to the midpoint of the trapezoid waist.
Example 3. In the known trapezoidal ABCD, ADC, E is the midpoint of AB, ED bisects ADC, and AD+BC=CD. Verification: ①EC⊥DE, ②EC divides ∠BCD equally.
Analysis: rotating △AED around point E makes A and B coincide, and point D falls on the extension line of CB, then △AED and △BEF are congruent, and it is easy to know that DE = Fe is from ∠2=∠F, then CD=CF, and a conclusion can be drawn according to the trinity property of isosceles triangle.
2.2.2 Issues related to the square
Rotate a triangle around the vertex by a certain angle, and the problem can be solved by graphic transformation.
In the square ABCD of example 4, m and n are on the sides of BC and CD, ∠ man = 45; Verification: MN=MB+ND.
Analysis: rotate △ clockwise around point A by 90 to coincide with △ABE, and you can get ∠ ean = 90, AE=AN, BE=DN, from ∠ man = 45, you can get ∠ EAM = ∠ man = 45, then △ AEM.
2.3 Using symmetry to add auxiliary lines
In the problem of sum and difference of line segments in a triangle, a triangle is often folded along the angle bisector with the help of the angle bisector to construct triangle congruence and carry out equivalent substitution.
Example 5. It is known that in the isosceles right triangle ACB, ∠ C = 90, and AD bisects ∠CAD. Proof: AB=AC+CD.
2 analysis: extend CD to e so that CE=CA=CB can be proved,
△ cam△ CEM; △ CBN △ cen, we can get: ME=MA, NE=NB, ∠ 1=∠A, ∠ 2 = ∠ b; So ∠ men = 90, using Pythagorean theorem: MN2=ME2+NE2=MA2=NB2. The above two situations are also called complementary to each other on the issue of adding auxiliary lines.
3 Other problems of adding auxiliary lines
3. 1 In the calculation and proof of the proportional line segment problem, when parallel lines are often drawn, one ratio in the conclusion is often retained, and then an intermediate ratio is connected with another ratio in the conclusion.
Example 7. In △ ABC, D is a point on AC, F is a point on CB extension line, and AD=BF, DF is given to AB in E, which proves that EF∶ED= AC∶BC.
Analysis: it is proved that the basic idea of this question is to add parallel lines, and the ratio EF∶ED can be retained when making parallel lines.
It is proved that 1: the intersection D is DM∨CF and AB is in M.
Then ef: ed = BF: DM.
AD∶DM= AC∶BC
AD = BF
∴EF∶ED= AC∶BC
Prove 2: FG∑AC passes through point F and extends the line from AB to G,
Then fg: ad = Fe: de.
AC∶BC= FG∶FB
AD = BF
∴EF∶ED= about 200 BC.
3.2 See the midpoint to lead to the neutral line, and use the nature of the neutral line.
Example 8. △ ABC, d is the midpoint of BC side and e is the midpoint of AD side. Connect BE, extend AC to point f, and prove that FC=2AF.
Proof 1: Analysis: Because D is the midpoint of BC side and E is the midpoint of AD side, it is easy to think of using the midline to solve the problem. As shown in figure 12, if the intersection point D is DG∑AC and BF intersects G, then G is the midpoint of BF, DG is the midline of △BFC, and FC = 2DG can be obtained; E is the midpoint of the AD side, DG∑AC, and it is easy to prove that DG=AF, so FC=2DG.
Proof 2: If D, DG∨BF crosses AC to G, and D is the midpoint of BC, then FG = GC; if E is the midpoint of AD and DG∨BF, then AF=FG, so AF=FG=GC, then FC=2DG can be obtained.
Example 9. Try to explain that the midpoints of the sides of a quadrilateral are connected in turn, and the resulting quadrilateral is a parallelogram.
It is known that in quadrilateral ABCD, E, F, G and H are the midpoint of AB, BC, CD and DA respectively. Explain that the quadrilateral EFGH is a parallelogram.
EF⊥AB can be known from the nature of trapezoid midline, and then it can be solved by the property of "three lines in one" of isosceles triangle. This problem can also be solved by prolonging the intersection of AF and BC and using the properties of the midline on the hypotenuse of a right triangle. In addition, by solving this problem, we can get two corresponding propositions: one is that the distance between the midpoint of a right-angled trapezoid and the two endpoints of another waist is equal, and the other is that the triangle area formed by the midpoint of any trapezoid and the two endpoints of another waist is equal to half of the trapezoid area. These two propositions can help us to examine problems in concrete problem solving. It is worth noting that the properties of triangle midline and trapezoid midline provide a new basis for explaining the parallel relationship and half-fold relationship of line segments in geometric problems, and create a new solution. Therefore, when dealing with geometric problems, we can think of the nature of the midline and construct the midline as an auxiliary line, which provides convenience for solving.