SAS means that both sides and their included angles are equal?
ASA is that two angles are equal to their sides?
AAS means that two sides are equal to the side next to them.
HL means that the right side and hypotenuse of a right triangle are equal.
As shown in the figure, it is known that in △ABC, ∠ ACB = ∠ ABC = 45, AB=AC, D is AC midpoint, AE⊥BD, and verification: ∠ADB=∠CDE. ?
Do AM⊥BC after point A, cross BD and point N,
∠ACB=∠ABC=45
∠BAM=∠ACB=45，∠CAE=∠ABD，AB=AC？
△ABN?△AEC，AN=CE？
∠MAC=∠C=45，AD=CD，AN=CE，
△ and△ ?△CED,
[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: The parallel line passing through C and AB intersects G, and FD=DG can be obtained by taking D as the midpoint of AC, and FG=GE can be obtained by 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, it is easy to get BH=HC= 1/4BE and ED∶EF=3∶4.
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, and the midpoint on the other side and the midline theorem of triangle can be used, which is 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:
(Figure 2) Construct a triangle OMC. Let DH⊥OC be in H, then ∠ 2 = 15 ∠ DCM = 15, then ⊿DMC≌⊿BOC and ∠ MCO = 60 DM = MC = OC =
∴∠dmo=360-60- 150 = 150
∴∠ 1=∠MOD= 15
So ∠ doc = ∠ dco = 75, DO=DC=AD=AB=AO.
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.
The conditions and conclusions of some geometric problems are scattered. By adding appropriate auxiliary lines, the scattered and "far away" elements in the drawing are gathered on the related drawings, so that they are relatively concentrated, which is convenient for comparison and establishment of relationships, thus finding 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, 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 to us by the questions, and find some implicit and regular information from them?
ADB =∠CDE