题目 #18822

(2) _{3}. (1)四边形BEDF是菱形, 证明过程如下: 由折叠的性质可知: EF垂直平分BD, ∴BE = DE, BF = DF, ∵ 四边形ABCD是矩形, ∴AD ∥BC,所以∠DEF = ∠BFE, ∵∠DFE = ∠BFE, ∴∠DEF = ∠DFE, ∴DE = DF,所以BE= DE = BF = DF, ∴四边形BEDF是菱形. (2)如图, BD与EF相交于点O,因为四边形ABCD是矩形, AB = 4, BC = 4√3, ∴BD =\(\sqrt{BC^{2}}\)+\(DC^{2}\)=\(\sqrt{4^{2}}\)+ （4√3）^{2}= 8,所以OB = OD =\(1 _{2}\)BD = 4, ∴AB = OB, ∵四边形BEDF是菱形, ∴∠BOE = 90∘,又∵ BE = BE, ∴Rt △ABE ≅Rt △OBE(HL), ∴AE = DE, 设OE = x, 则BE = DE等于AD −AE = 4√3 −x, 在Rt △ OBE中,\(OE^{2}\)+\(OB^{2}\)=\(BE^{2}\), ∴\(x^{2}\)+\(4^{2}\)= （4√3 −x）^{2},所以8√3x = 48 −16, ∴x = 4√3 8\(\sqrt{3 _{3}}\), ∴EF = 2OE = _{3}, 故答案是: 8\(\sqrt{3_{3}}\). 13.(1)见解析; (2)∠BAC = 110∘. (1)证明: 连接AD, ∵D是BC的中点, AB = AC, ∴AD平分∠BAC, ∵DE ⊥AB, DF ⊥AC所以DE = DF; (2)∵DE ⊥AB, ∴∠BED = 90∘, ∵∠BDE = 55∘, ∴∠B = 35∘∴∠C = 35∘,所以∠BAC = 110∘. 14.证明见解析 ∵四边形ABCD是平行四边形，∴AB ∥CD，∴∠BAE = ∠CFE，∠ABE = ∠FCE.∵E为BC的中点，∴EB = EC，∴△ABE ≌△FCE，∴AB = CF.又∵AB ∥CF，∴四边形ABFC是平行四边形.∵AD = BC，AD = AF，∴ BC = AF，∴四边形ABFC是矩形. (1)见详解