On the website in the lower right corner of the course guide. Go register and download. The first stage:11.1~1.2 (1) consolidate the basis of the test questions. First, choose carefully and make the final decision. The condition in 1. d. 2. d. 3. c. 4. b can draw two triangles; The conditions in d can draw countless triangles. Second, think carefully and skillfully, make the finishing point 7.4.8. CD = CB or ∠ DAC = ∠ DAC = ∠ BAC.9.65.10.22 .. Tip: If △ ABC △ dcb is proved first, then ∞. Victory is in sight 1 1. Solution: According to the meaning of the question ∠ b = ∠ c = 30. ∴∠ BFC = ∠ A+∠ b = 80, ∴∠ BOC. That is BC = ef, and ∵ AB = DE, ∴△ ABC △ def (SAS). ∴ ∠ A = ∠ D. 13。 Proof: OA = Ob, OC = OD, AC. (2) Select "If ①, ③, ②" to prove it. Proof: ∫be∫af, ∴ AFD = ∠ BEC. And ∵∠ A = ∠ B, AD = BC, ∞. ∴∠ ABC =∠ CBF = 90。 In △ABE and △CBF, ∴△ Abe △ CBF (SAS). ∴ AE = cf. (2) From the meaning of the question, △ABC and △EBF are isosceles. ∴∠ AEB =∠ CAE +∠ ACB = 30+45 = 75。 From (1), △ Abe △ CBF, ∴∠CFB =∞. ∠ ADB = 100，∴∠ AEC = ∠ ADB。 ∠∠Bad+∠CAE = 80，∠ ACE+∠ CAE = ∠ CED。 △ a ′ df △ CB ′ e. Select △ aa ′ e △ c ′ cf for explanation. ad = cb，∠d =∠b = 90°，AB = CD，∴△ab = CD?△。 ∴△aa′e?△c′cf(asa)。 4.( 1)≈A+∠APB = 90，∠ APB+∠ QPC = 90，∞。 Then PC = BP-BC = 2 = AB. Then △ BAP △ CPQ (ASA) and ∴ PA = PQ. To sum up, when BP = 3 or BP = 7, PA = PQ. The third stage: the first chapter XI comprehensive test questions (1) First, carefully selected, the final tone is1.d.2.b.3.c.4.c.5.a.6.c.7.c.8.b.9.c./. Make the finishing touch11.27.12.60.13.65438 The answer is not unique. Such as eh = be or △AEB = ce or ah = bc. 15. Vertical.16.100.17.10.18. (8, 6) Third, victory is in sight. ∴∠ b =∠ c.20. △ a1b1c1and △∠BCD are not necessarily congruent, and the figure is omitted. 21.△ af ⊥ df △ AE ⊥ be, reason: △ AC is equally divided. MF, ∫ab∨CD, ∴ B = ∠B=∠C. In △BEM and △CFM, be = CF, ∠B=∠C, BM = CM, ∴△BEM?. ∴∠ Asian Development Bank =∠ ADC. ∫AE is the angular bisector, ∴∠ BAE =∠ CAE, AD = AD, ∴△ABD≌△ACD(ASA), ∴.∴△ABE≌△ACE(SAS ∴ AP = AQ。 (2)∫△ABP?△qca，∴∠ P = ∠ CAQ。 And ∵∠ P+∠ Pad = 90, ∴∠.∴ BD = 5cm. And ∵ PC = BC-BP, BC = 8cm, ∴ PC = 8-3 = 5cm, ∴ PC = BD. And ∵∠ B = ∠ C, ∴△ BP. Figure 3 is not valid. Prove Figure ②. Extend DC to point K to make CK = AE and connect BK, then △ BAE △ BCK, ∴ BE = BK, ∠ Abe = ∠ KBC. ∠ FBE = 60，∞。 ∴ KC+CF = EF, which means AE+CF = EF. Figure ③ does not hold, and the relationship among AE, CF and EF is AE-CF = EF. Chapter 11 Comprehensive Test Questions (2) First, carefully select and finally decide:/kloc-0 1.C.2.A.3.C.4 Make the finishing touch 1 1. ∠ DBE，AC. 12.30. 13。 The answer is not unique, such as ∠ b = ∠ d14. The answer is not unique, such as RT △ choose two of them. 15.145.16.78.17.7.18.① ② ④. Third, victory is in sight 19. BC = BD。 ∴△ Abbe△ ?△abf. 20. consistency Press fold, △ BDE △ BDC. ∴ DE = DC，∠e =∠c = 90°。 ab = dc，∴ AB。 Proof: diagonal AC equal division ∠ bad. It is proved that point C is perpendicular to AB and AD, and the vertical feet are points E, F, ∫≈ADC+∠B = 180, ∠ADC+∞△CDF = 65438+ respectively. (2)FC = EA； (3) Prompt: use SAS CF⊥BD △ Abe △ CDF.23. ∫∠ B = 90, ED⊥AC at point D, BE = DE, ∴AE to share ∠BF⊥AC, ∴∠EAD =(2)① ∠∠BAC = 90°，∴ ∠ DAF = ∠ BAC，∴∠ DAB。 Intercept OE = of on OA and OB, take any point C on OP and connect CE and CF, then △ COE △ COF; (2) Intercept am = AE on AC, followed by FM, where AD is the bisector of ∠BAC, ∠ EAF = ∠ MAF, and ∵ AF = AF, ∴△AEF≌△AMF, ∴ EF = ∴. And ∵ CF = CF, ∞. (2) ① α+β = 180. Reason: ∫∠BAC =∠DAE, ∴∠ BAC-∠ DAC = ∠ DAE-∠ DAC, that is ∠ bad. ∴ α+β = 180.② When point D is on ray BC, α+β = 180, and when point D is on the reflection extension line of ray BC, α = β. ), (8,8), (8,6) or (8,6).