The most common way to prove the equality of line segments (or angles) by congruent triangles is to explore that the verified line segments (or angles) are in a pair of provable congruent triangles. AB and DC in this problem belong to two pairs of triangles △ABE and △CDE and △ABC and △ DBC respectively. It can be proved by analysis that △ Abe △ CDE.
Syndrome known ∠ 1=∠2,
∠ABC=∠DCB, and
∠EBC=∠ABC-∠ 1，
∠ECB=∠DCB-∠2，
So ∠ EBC = ∠ ECB. In ...
In △ABC and △BCD,
∠ABC=∠BCD，
∠EBC =∠ European Central Bank, BC=BC,
So △ ABC△ DCB (ASA),
So ab = CD.
It shows that line segments AB and CD also belong to two (actually) congruent △ABE and △DCE, so it can also be directly proved that these two triangles are congruent.
Example 2 is shown in Figure 2-2. △ ABC is an isosceles triangle, D and E are points on the extension line of waist AB and AC respectively, and BD=CE connects d E and BC. Verification: GD = GE.
Graphically, GE and GD belong to two obviously unequal triangles: △GEC and △ GBD. At this time, it is necessary to use the existing equivalent conditions in these two triangles and combine the known auxiliary lines to construct the congruent triangles. There is more than one method, and the following proof is one of them.
It is proved that E is EF‖AB and the extension line passing through BC is F. At △GBD and △GEF, ∠BGD=∠EGF (diagonal), ①.
∠B=∠F (the internal dislocation angles of two parallel straight lines are the same). ②
And ∠B=∠ACB=∠ECF=∠F, so △ECF is an isosceles triangle, so EC = ef. And because EC=BD, so
BD=EF。 ③
Pass ①, ②, ③
△ GBD△ Global Environment Facility (AAS),
So GD = ge.
It shows that there can be more than one method to construct congruent triangles by adding auxiliary lines appropriately. There are at least two methods in this topic:
(1) cross d as DF‖AC and BC as f, and you can prove △ gfd △ GCE in the same way (Figure 2-3).
(2) if d is used as df ⊥ BC in f; If E is EH⊥BC and the extension line of BC is H, it can be proved that △ GFD △ GEH (Figure 2-4).
After finishing a problem, think about whether there are other proof methods and compare which one is better, which is of great benefit to developing thinking and exercising ability.
Example 3 is shown in Figure 2-5. In the equilateral triangle ABC, AE=CD, AD, BE intersect at point P, and BQ⊥AD intersect at point Q. Proof: BP=2PQ ..
The analysis first shows that BP and PQ are in Rt△BPQ, as long as it is proved that ∠BPQ = 60 (or ∠ PBQ = 30). But ∠ BPQ is the outer corner of △ABP, so ∠ BPQ = ∠ PAB+∠ PBA.
Proved in △ABE and △CAD,
∠EAB=∠DCA=60，AB=CA，AE=CD，
therefore
△ABE?△CAD(SAS)，
So ∠ABE=∠CAD ..
Because ∠BPQ is the outer corner of △ABP, so
∠BPQ=∠PAB+PBA=∠PAB+∠CAD=60。
In Rt△BQP, ∠ bpq = 60 and ∠ pbq = 30, so BP=2PQ (in RT △ bqp, the opposite side of 30 is equal to half of the hypotenuse).
It shows that finding or constructing congruent triangles is the key to prove the problem with congruent triangles. Therefore, we often start from finding the equality of the corresponding elements in two triangles, and gradually find or "make up" the condition of triangle congruence through reasoning. For example, after analyzing the problem and proving ∠△ABP=∠CAD, we will focus on △ABE and △CAD. Here, we can use geometric intuition to inspire us to find promising consistency.
Example 4 is shown in Figure 2-6. ∠ A = 90, AB=AC, M is the midpoint of AC side, AD⊥BM intersects BC in D, and BM intersects E. Verification:
∠AMB=∠DMC。
Analysis 1 From the graphic observation, the two triangles △AME and △DMC where ∠AME and △ DMC are located are obviously different, but there are other equivalent elements in these two triangles: AM = MC. It is ideal if we can construct new equivalent elements in existing triangles to form congruent triangles by using known conditions. Because ∠ C = 45. Then, in △AGM, ∠ GAM = 45 = ∠ C. Combined with ∠AMB=∠DMC in verification (this can't be regarded as known, of course, but it can be regarded as known in analysis in order to find ideas), we can assert that △AGM "should" and △CDM are congruent. To do this, we only need to find a set of equal sides in these two triangles. The graph and conditions inspire us to consider proving △ AGB △ CDA.
It is proved that 1 is the bisector AG of ∠BAC, which intersects BM in G, and it is in △AGB and △CDA, because
AB=CA，∠BAG=∠ACD=45，
∠ABG=90 -∠AMB，①
∠MAD=90 -∠EAB。 ②
In Rt△MAB, AE⊥BM, so ∠ AMB = ∠ EAB. From ①, ②, ∠ABG=∠MAD, so
△AGB?△ADC(ASA)，
So ag = CD.
In △AMG and △CMD, there are also
AM=MC，∠GAM=∠DCM=45，
So △ AMG △ cmd,
So ∠ AMB = ∠ DMC.
Analysis 2 is shown in Figure 2-7. Note that in Rt△ABM, you get ∠MAE=∠MBA from AE⊥BM. If you extend AE, take C as CF⊥AC and extend AE to F, you can form Rt△ABM≌Rt△ACF, so ∠ AMB =
Evidence 2 introduces the auxiliary line described in analysis 2. In Rt△ABM and Rt△CAF, ∠ABM=∠CAF, AB=AC, and
∠BAM=∠ACF=90，
therefore
Rt△ABM≌Rt△CAF(ASA)，
therefore
∠AMB=∠F, AM = See ①.
In △MCD and △FCD, FC=AM=MC (because M is the midpoint of AC). Because ∠ ACF = 90, ∠ ACB = 45
∠FCD=∠MCD=45，CD=CD，
So △ fcd △ MCD (SAS)
So ∠ f = ∠ DMC ②
By ①, ② ∠ AMB = ∠ DMC.
It shows that the ideas of these two proofs are complicated. The result of adding auxiliary lines is to produce two pairs of congruent triangles, the first pair of congruent triangles produces some equivalent elements, paving the way for the second pair of congruent triangles. The first pair of congruent triangles "transferred" a corner to the second pair of congruent triangles, so the problem was finally solved. For some complicated problems, we should adopt a circuitous method to create a congruent triangles according to the situation, so as to make the conditions more equal, move the corner (or edge) to be proved out of the "dead corner" and finally solve the problem.
Example 5 is shown in Figure 2-8. In a square ABCD, take any point Q, connect AQ, cross D as DP⊥AQ, cross R as AQ, cross P as BC, and the diagonal intersection of the square is O, connect OP and OQ. Verification: ⊥ Observatory.
If you want to prove OP ∠ OQ, you can prove ∠ COP+∠ COQ = 90. But ∠ COQ+∠ QOD = 90, so you only need to prove ∠COP=∠DOQ. This boils down to proving △COP∞.
In the square ABCD, because AQ⊥DP, there is ∠ RDQ = ∠ QAD in Rt△ADQ and Rt△RDQ. So there are Rt△ADQ and Rt△DCP.
AD=DC，∠ADQ=∠DCP=90，
∠QAD=∠PDC，
therefore
△ADQ?△DCP(ASA)，DQ=CP。
At △DOQ and △COP,
DO=CO，∠ODQ=∠OCP=45，
therefore
△DOQ?△COP(SAS)，∠DOQ=∠COP。
therefore
POQ=∠COP+∠COQ=∠DOQ+∠COQ
=∠COD=90，
This is ⊥ Ou Ke.
It shows that (1) using the special properties of special graphs, we can often find useful conditions, such as the diagonal lines of a square are perpendicular to each other, the diagonal lines make an angle of 45 with the edge, OA=OB=OC=OD, etc. Used to derive congruent triangles.
(2) The congruences of two triangles are equal to the corresponding elements, which are mutually causal. This is the basic skill of using congruent triangles to prove the problem.
Example 6 is shown in Figure 2-9. In the known square ABCD, m is the midpoint of CD, e is the upper point of MC, ∠ BAE = 2 ∠ DAM. Verification: AE = BC+CE.
There are two basic methods to prove that a line segment is equal to the sum of two line segments:
(1) A line segment is "constructed" by adding auxiliary lines to make it the sum of two line segments in verification (BC+CE), and it is proved that the constructed line segment is equal to the line segment in verification.
(2) By adding auxiliary lines, firstly intercept the line segment equal to one segment (such as BC) in the long line segment (AE) in verification, and then prove that the cut part is equal to another segment (CE) in the line segment. We use (1) to prove it.
If we extend AB to F and make BF=CE, we can know from the property of square.
AF=AB+BF=BC+CE。
Let's use congruent triangles to prove AE = AF. To this end, connect the EF intersection BC to G. Because the diagonal ∠BGF=∠CGE, so
Rt△BGF≌Rt△CGE(AAS)，
therefore
therefore
Rt△ABG≌Rt△ADM(SAS)，
therefore
G leads GH⊥AE to H. Since AG is the bisector of ∠EAF, GB=GH, Rt△GBF≌Rt△GHE(HL), so
∠F=∠HEG，
AF=AE (triangles with equal base angles are isosceles triangles),
That is AE = BC+ce.
It shows that we can also prove the conclusion according to the method of analysis (2). For this reason, we can make the bisector AG of ∠BAE ∠ intersect BC in G, and then make GH⊥AE in H. By proving △ ABG △ AHG, we can know that AB = AH = BC. Let's try to prove that he =CE. Please prove it yourself.
Exercise 10
1. as shown in figure 2- 10, AD, EF and BC intersect at point o, AO=OD, BO=OC, EO = of. Verification: △ AEB △ DFC.
2. As shown in Figure 2- 1 1. In the regular triangle ABC, P, Q and R are the midpoint of AB, AC and BC respectively, M is any point on BC (different from R), and △PMS is a regular triangle. Verification: RM = QS.
3. As shown in Figure 2- 12. P is any point on the diagonal BD of pf⊥dc ABCD, PE ⊥ BC square. Verification: AP ⊥ ef.
4. As shown in Figure 2- 13, the high AD and BE of △ ABC intersect at H and BH = AC. Verification: ∠ BCH = ∠ ABC.
5. As shown in Figure 2- 14. In a square ABCD, p and q are BC, the points on the side of CD, ∠ PAQ = 45. Verification: PQ = Pb+DQ.
6. As shown in Figure 2- 15, the vertex A passing through △ABC is perpendicular to the bisector of two base angles ∠B and ∠C respectively, with AD⊥BD in D and AE⊥CE in E. Verification: ED ‖ BC.