Example 1 known: as shown in figure 1- 1, in quadrilateral ABCD, BC > Ba, AD=CD, BD bisects ∠ABC, which is verified as ∠∠ A+∠ C =1. Analysis: Because right angles are equal to. Therefore, the key to solve the problem is to construct an isosceles triangle, which can be achieved by "giving up the strengths to make up for the weaknesses". It is proved that BE=AB, connect DE on BC, and then take the middle point m of EC, connect DM∶ab = be and ∶BD to divide equally ∠ABC a d∴△Abd =∞△EBD in. Ab = be ∠ Abd = ∠ EBD BD = BDB EMC ∴△ Abd △ EBD (SAS) as shown in figure 1- 1 ∴ AD = ED ∠A = ∠BED,. ∴ ed = DC ∴∠ c = ∠ dec ∴∠ a+∠ c = ∠ bed+∠ dec =180 Example 2 Known: As shown in Figure 2-2, AE//. Verification: AB=AE+BC analysis 1: It needs to be proved that AB=AE+BC observed that AD and BD are bisectors of angles, so DAED can be folded along A, so auxiliary lines need to be added to AB to intercept BF=BC. Only by deducing AF=AE can the problem be solved, so how to deduce AF=AE becomes the key to solve the problem. Because the sum of internal angles of DAED, DADB and DBD is 180, while EDC = 180, and because AE//BE and E+C = 180, EAB+CBA = 180, and AD and BD are straight angles. If we can deduce 7=8, then we only need to deduce DAED≌DAFD, then we can deduce AE = AF. Because BC=BF, 1=2, BD is the public side, so we can deduce DBFD≌DBCD, so it is 5 = 6. Because 5+7 = 90. Prove 1: intercept BF=BC on AB and connect DF. Bd is the bisector of ABC, ∴ 1=2 is in DBDF and DBDC (male * * * side) ∴DBDF≌DBDC(SAS) as shown in Figure 2-2 ∴5=6 (congruent triangles corresponding angles are equal) ∴ 3+8+E =. BC ∴ E+C = 180 (two lines are parallel and complementary) and ∵ EDC =180 ∴1+2+3+4 =180 ∴ AD is EAB. In D and DAFD, ∴DAED≌DAFD (ASA) ∴AE=AF (congruent triangles's corresponding sides are equal) ∫af+FB = ab ∴ae=fb=ae+bc=ab=ae+bc Analysis 2: Extend the extension line of BC to AD at F, and prove it. To prove CF=AE, we only need to deduce that DAED≌DFCD of two triangles containing CF and AE is 5=6, AE//BC, and then we can deduce 3 = F. If we want to deduce AD=FD, it becomes the key to solve the problem, because the sum of the internal angles of the quadrilateral AECB is equal to 360, and E+BCE = 180. And because AD and BD are the bisectors of EAB and CBA, 1+4 = 90 can be deduced, so ADB=90 and EDB=90. From this, they can infer that DABD≌DFBD is based on ASA by observing the graph, thus deducing AD=FD, and the idea is formed. Prove 2: As shown in Figure 2-3, the sum of the internal angles of BC, AD and F triangles extended in DAED, DADB and DBDC is 540 (triangle internal angle sum theorem); > EDC = 180 (rectangular definition) ∴ E+C+EAB+ABC = 180. BC ∴ (two lines are parallel and their internal angles are complementary) ∴ 3+4+ 1+2 = 180 ∴AD and BD are the bisectors of EAB and ABC ∴3=4, respectively. 1=2 (angle bisector definition) ∴ 1+4 = 90 ∴ ADB = 90 (triangle interior angle sum theorem) ∴ BDF = 90 in DADB and DBDF ∴ DADB ≌ DBDF (. The corresponding angles are equal) as shown in Figure 2-3 ∴ DAED ≌ DFCD ∴ AE = FC ∴ BF = BC+FC ∴ BF = BC+AE ∴ AB = AE+BC Example 3 is known: as shown in Figure 3- 1. Proof: Method 1: Truncation intercepts AE = AC on AB and connects ED. A: AD is divided into ∠ BAC ∴ Bad = ∠ DAC in △ADE and △ADC, and E AE = AC ∠EAD= ∠DAC B D C AD = AD as shown in Figure 3-1∴△ ade. Be > BD-DE (the difference between two sides of a triangle is smaller than the third side), that is, AB-AE > BD-DC ∴ AB-AC > BD-DC (equivalent substitution) Method 2: Complement method extends AC to point E, so that AE = AB and connects Ade: AD ∴ BAC ∞. AB = AEC ∠ bad = ∠ DAC B D E AD = AD ∴△ Ade△ ADC (SAS) ∴ DB = D E as shown in Figure 3-2, at △ABD, EC >DE —DC (the difference between two sides of a triangle is less than the third side), that is, AE-AC > de-DC. It is proved that the first method (complement method) extends AC to E to make DC=CE, then ∠CDE =∞△aed, as shown in Figure 4-2 ∠ ACB = 2 ∠ E. ∴AB=AE. Figure 4-3 shows AE=AC+CE=AC+DC, \. In △∠AFD=∠ACD △ACD，△AFD?△