Mathematics in the second day of the May 4th Movement.

Example 1: As shown in figure 1- 1: D and E are two points in △ABC, which proves that AB+AC > BD+DE+CE.

Proof: (method 1) extend both sides of DE to AB and AC respectively. In m, n,

In △AM+AN, AM+AN? >? MD+DE+NE； ( 1)

At △BDM, MB+MD > BD; (2)

At △CEN, cn+ne > ce；; (3)

From (1)+(2)+(3):

? AM+AN+m b+ MD+CN+NE > MD+DE+NE+BD+CE

∴AB+AC>BD+DE+EC？

(Method 2:) As shown in figure 1-2,? Extend BD delivery? AC in f, extending CE to BF in g,

In △ABF and △GFC and △GDE, there are:

AB+AF >？ BD+DG+GF？ (The sum of two sides of a triangle is greater than the third side) (1)

Gf+fc > ge+ce (same as above) .......................................... (2)

DG+GE > DE (ditto) ...................... (3)

From (1)+(2)+(3):

a b+ AF+GF+FC+DG+GE > BD+DG+GF+GE+CE+DE

∴AB+AC>BD+DE+EC。

Second, when the external angle of a triangle is larger than any internal angle that is not adjacent to it, if it cannot be directly proved, the triangle can be constructed by connecting two points or extending an edge, so that the verified position of the big angle is in the external angle of the triangle and the small angle is in the internal angle of the triangle, and then the external angle theorem can be used:

For example, as shown in Figure 2- 1, it is known that d is any point in △ABC and verified as ∠ BDC > ∠ BAC.

Analysis: Because ∠BDC and ∠BAC are not in the same triangle, there is no direct connection, so a new triangle can be constructed by adding auxiliary lines appropriately, so that ∠BDC is in the outer corner and ∠BAC is in the inner corner;

Prove 1: extend the intersection of BD and AC to point e, when ∠BDC is the outer corner of △EDC.

∴∠ BDC >∠ DEC, in the same way ∠ DEC >∠ BAC, ∴∠ BDC >∠ BAC.

Method 2: Connect AD and extend the delivery of BC to F ..

∫∠BDF is the outer corner of △ABD.

∴∠ BDF >∠ bad, in the same way, ∠ CDF >∠ CAD.

∴∠BDF+∠CDF>∠BAD+∠CAD

Namely: ∠ BDC > ∠ BAC.

Note: When proving inequality with triangle exterior angle theorem, we usually put the big angle on the exterior angle position of triangle and the small angle on the interior angle position of triangle, and then prove it with inequality properties.

Third, there is an angle bisector. Usually, equal line segments are intercepted on both sides of the angle to construct an congruent triangle, such as:

For example, as shown in Figure 3- 1, it is known that AD is the center line of △ABC, and ∠ 1 = ∠ 2, ∠ 3 = ∠ 4, verification: be+cf > ef.

Analysis: to prove be+cf > ef? We can prove that BE, CF, EF must be moved to the same triangle by using the triangle trilateral relation theorem. From the known ∠ 1 = ∠ 2, ∠ 3 = ∠ 4, we can intercept equal line segments on both sides of the angle and move en, FN, EF to the same triangle by using the congruent corresponding edges of the triangle.

It is proved that if DN = DB is intercepted on DA, NE and NF are connected, and DN = DC,

At △DBE and △DNE:

∵

∴△DBE≌△DNE？ (SAS)

∴ be = ne (the corresponding sides of congruent triangles are equal)

Similarly: cf = nf

In △EFN, en+fn > ef (the sum of two sides of a triangle is greater than the third side).

∴BE+CF>EF。

Note: When there is an angular bisector in the proof, it is often possible to construct a congruent triangles by intercepting line segments with equal sides of the angle, and then use the properties of congruent triangles to obtain the equality of the corresponding elements.

4. When the midpoint of a line segment is the endpoint, it is often extended and doubled to form an congruent triangle.

For example, as shown in figure 4- 1: AD is the center line of △ABC, and ∠ 1 = ∠ 2, ∠ 3 = ∠ 4, and verification: be+cf > ef.

Proof: extend ED to m so that DM=DE, connect?

Cm, if. In △BDE and △CDM,

∵

∴△BDE≌△CDM？ (SAS)

∵∠ 1 =∠ 2, ∠ 3 =∠ 4? (known)

? ∠ 1+∠ 2+∠ 3+∠ 4 = 180 (definition of right angle)

∴∠3+∠2=90

Namely: ∠ EDF = 90.

∴∠FDM=∠EDF？ =90

In △EDF and △MDF.

∵

∴△EDF≌△MDF？ (SAS)

∴EF=MF？ (The corresponding sides of congruent triangles are equal)

∵ in △CMF, cf+cm > MF (the sum of two sides of a triangle is greater than the third side)

∴BE+CF>EF

Note: FD can also be doubled in the above problems, and the proof method is the same as above.

Note: When there is a line segment whose endpoint is the midpoint of the line segment, the line segment can be extended and folded in half to form a congruent triangles, so that the scattered conditions in the question can be concentrated.

5. When there is a triangular midline, the double midline is often extended to form the congruent triangles.

For example, as shown in Figure 5- 1: AD is the center line of △ABC, and verification: AB+AC > 2AD.

Analysis: In order to prove AB+AC > 2AD, the diagram shows:? AB+BD > AD, AC+CD > AD, so there is AB+AC+? BD+CD > AD+AD = 2AD, the BD+CD on the left has more conclusions to prove, so we can't prove this problem directly, but 2AD thought of constructing 2AD, that is, doubling the midline and transferring the line segments to the same triangle.

Proof: extend AD to E, make DE=AD, connect BE, AE = 2AD.

∵AD is the center line of △ABC? (known)

∴BD=CD？ (midline definition)

In △ACD and △EBD,

∴△ACD≌△EBD？ (SAS)

∴ Be = Ca (the corresponding sides of congruent triangles are equal)

∵ in △ABE, there are: AB+BE > AE (the sum of two sides of a triangle is greater than the third side).

∴AB+AC>2AD。

(The midline is usually extended and folded to form an congruent triangle)

Exercise: It is known that △ABC and AD are the median lines on the side of BC, and an isosceles right triangle is made with AB and AC as right angles, as shown in Figure 5-2. Verify ef = 2ad. ?

Six, interception, complement each other as an auxiliary line.

For example, as shown in Figure 6- 1, at △AB>AC, AB>AC, ∠ 1 = ∠ 2, p is any point of AD.

Proof: AB-AC > PB-PC.

Analysis: to prove: AB-AC > PB-PC, I want to prove it with the triangle trilateral relation theorem. Because what I want to prove is the difference between line segments, and the difference between two sides is smaller than the third side, so I want to construct the third side AB-AC. Therefore, I can intercept an an on AB and get AB-AC = BN. If PN is connected again, PC = pn, and in △PNB, Pb-pn < bn,

Namely: ab-AC > Pb-PC.

Proof: (truncation method)

Intercept an = AC connection PN on AB? ,? At △APN and △APC,

∵

∴△APN≌△APC？ (SAS)

∴PC=PN？ (The corresponding sides of congruent triangles are equal)

∵ In △BPN, is there? PB-PNPB-PC。

Seven, extend the known edge structure triangle:

For example, as shown in Figure 7- 1, it is known that AC = BD and AD⊥AC is in A? , BC⊥BD in B,? Verification: ad = BC

Analysis: Want to prove? AD = BC. It is proved that the triangle containing AD and BC is congruent. There are several schemes: △ADC and △BCD, △AOD and △BOC, △ABD and △BAC. However, according to the existing conditions, it is impossible to prove that the congruence and the difference angle are equal, so we can try to make a new angle to make it the common angle of two triangles.

It is proved that DA and CB are extended separately, and their extensions intersect at point e,

? ∵AD⊥AC BC⊥BD (known)

? ∴∠CAE=∠DBE？ = 90 (vertical definition)

In delta delta δ△DBE and delta delta delta δ△CAE

? ∵

? ∴△DBE≌△CAE

∴ED=EC？ EB=EA？ (The corresponding sides of congruent triangles are equal)

∴ED-EA=EC-EB？

Namely: ad = BC.

(When the conditions are insufficient, you can obtain new conditions by adding auxiliary lines to create conditions for proof. )

Eight? Connect the diagonals of quadrilateral, and solve the quadrilateral problem by transforming quadrilateral into triangle.

For example, as shown in Figure 8- 1:AB∨CD, AD∨BC Verification: AB=CD.

Analysis: The graph is a quadrilateral, and we have only learned about triangles, so we must transform it into triangles to solve it.

Proof: Connect AC (or BD)

? ∵AB∨CD？ AD ∨ BC? (known)

? ∴∠ 1=∠2,∠3=∠4? (Two straight lines are parallel and the internal dislocation angles are equal)

In △ABC and △CDA

? ∵?

? ∴△ABC≌△CDA？ (ASA)

? ∴ AB = CD (the corresponding sides of congruent triangles are equal)

Nine, when there is a line segment perpendicular to the bisector, this line segment is usually extended.

For example, as shown in Figure 9- 1, in Rt△ABC, AB = AC, ∠ BAC = 90, ∠ 1 = ∠ 2, and CE⊥BD is longer than E? . Verification: BD = 2ce?

Analysis: To prove BD = 2CE, I want to construct a bisector with 2CE and CE perpendicular to ∠ABC. what do you think? To extend it.

Proof: extend the intersection of BA and CE at point f respectively.

? ∵BE⊥CF？ (known)

? ∴∠BEF=∠BEC=90？ (definition of vertical)

At △BEF and △BEC,

∵?

∴△BEF≌△BEC(ASA)∴CE=FE=CF？ (The corresponding sides of congruent triangles are equal)

? ∵∠BAC=90？ BE⊥CF？ (known)?

∴∠BAC=∠CAF=90？ ∠ 1+∠BDA = 90∠ 1+∠BFC = 90

? ∴∠BDA=∠BFC

In △ABD and △ACF

∴△ABD≌△ACF？ (AAS)∴BD=CF？ (The corresponding sides of congruent triangles are equal) ∴ BD = 2ce.

10. Connect known points to construct congruent triangles.

For example, it is known as shown in figure10-1; AC and BD intersect at point o, AB = DC, AC = BD. Verification: ∠ A = ∠ D.

Analysis: To prove ∠ A = ∠ D, we can prove that their triangles △ABO and △DCO are congruent, but it is difficult to prove their congruence only under the conditions of AB = DC and antipodal angle, so we only need to find other triangles, from AB = DC and AC = BD. If BC is connected, then △ AB=DC and △

Proof: connect BC, in △ABC and △DCB.

∵?

∴△ABC≌△DCB？ (SSS)

∴∠A=∠D？ (The corresponding sides of congruent triangles are equal)

Eleven, take the midpoint of the line segment and construct three congruent figures.

For example, as shown in figure11:ab = DC, ∠ A = ∠A=∠D? Verification: ∠ ABC = ∠ DCB.

Analysis: From AB = DC, ∠ A = ∠ D, it is considered that if the midpoint n of AD connects NB and NC, then the SAS axiom has △ ABN △ DCN, so BN = CN, ∠ ABN = ∠ DCN. We only need to prove that ∠ NBC = ∠ NCB, and then take the midpoint m of BC and connect MN, then the SSS axiom has △ NBM △ NCM, so ∠ NBC = ∠ NCB. The problem is proved.

Proof: take the midpoint n and m of AD and BC to connect NB, NM and NC. Then AN=DN, BM=CM, and in △ABN and △DCN, ∫△ABN?△DCN? (SAS)

? ∴∠ABN=∠DCN？ NB=NC？ An congruent triangle has equal sides and angles.

At △NBM and △NCM,

∵

∴△NMB≌△NCM,(SSS)？ ∴∠NBC=∠NCB？ (congruent triangles has equal corresponding angles) ∴∠ NBC+∠ ABN? =∠NCB+∠DCN

That is ∠ ABC = ∠ dcb.

Ingeniously calculate the proportion of line segments in triangle

Example 1. As shown in figure 1, in △ABC, BD: DC = 1: 3, AE: ED = 2: 3, and find AF: FC.

Solution: point d DG//AC, point g BF? So DG: fc = BD: BC

Because BD: DC = 1: 3.

So BD: BC = 1: 4.

That is, DG: FC = 1: 4, FC = 4DG because DG: AF = DE: AE?

And because AE: ed = 2: 3.

So DG: AF = 3: 2, so AF: FC =: 4dg = 1: 6.

Example 2. As shown in figure 2, BC = CD, AF = FC, and find EF: FD.

Solution: If CG//DE passes through point C and AB passes through point G, there is ef: GC = AF: AC.

Because AF = FC, AF: AC = 1: 2 is EF: GC = 1: 2.

Because CG: de = BC: BD?

Because BC = CD, BC: BD = 1: 2? CG: DE = 1: 2, which means DE = 2GC.

Because FD = ed-ef =, ef: FD =

Summary: In the above two cases, the auxiliary line is made at the intersection of two known line segments under the condition of "known", and the auxiliary line is parallel to the line segment appearing in the conclusion. Please look at two more cases and let us feel the mystery!

Example 3. As shown in figure 3, BD: DC = 1: 3, AE: EB = 2: 3, and find AF: FD.

Solution: Pass point B as BG//AD and cross CE extension line at point G, so df: BG = CD: CB.

Because BD: DC = 1: 3.

So CD: CB = 3: 4

That is, df: BG = 3: 4.

Because af: BG = AE: EB?

Because AE: EB = 2: 3, AF: BG = 2: 3.

So af: df =

Example 4. As shown in Figure 4, BD: DC = 1: 3, AF = FD, and find EF: FC.

Figure 4

Solution: Take point D as DG//CE, and AB crosses at point G..

So ef: DG = af: ad