导读:c=3B在△ABC中,A、B、C的对边分别为a、b、c,当3∠A+4∠B=180▫,c=3, b=2, 那么a = ____________B=2CA解题过程:① 过B点做线段BD,与AC延长线交于D,并使BD = AB② 连接
c=3
B
在△ABC中,A、B、C的对边分别为a、b、c,当3∠A+4∠B=180▫,c=3, b=2, 那么a = ____________
B=2
C
A
解题过程:
① 过B点做线段BD,与AC延长线交于D,并使BD = AB
② 连接AD,在AD上截取一点E,连接BE,使BE = DE
设∠A为ɑ,∵ BD = AB,BE = DE,∴ ∠A= ∠D = ∠DBE = ɑ
③ 过B点做线段BF,与AD交于F,并使∠CBF = ∠CBA
设∠CBA为β,则∠CBF = ∠CBA = β,∠ABF= 2β
B
(完成辅助线后见下图)
④ 在∆ABD中,内角和 = 180▫
∴ ∠D+∠DBE+∠EBF+∠ABF+∠A = 180˚
∴ 3ɑ+2β+∠EBF =180˚
∵ 3∠A+4∠B = 180˚,且∠A= ɑ,∠B = β
∴∠EBF = 2β
⑤ 在∆BDF中,∠DBF = ∠DBE+∠EBF = ɑ+2β(见④的结论)
∠BFD = ∠ABF+∠A = 2β+ɑ(外角定理),∴∠DBF =∠BFD
∴ DF = DB = AB = c = 3
⑥ 在AB上截取一点H,使BH = BF,易证∆BFC ≌ ∆BHC,∴BH = BF,
∴∠BCF = ∠BCH,∠BFC = ∠BHC
∵ ∠BCF = ∠ABC+∠A = ɑ+β, ∴∠FCH = ∠BCF+∠BCH = 2∠BCF = 2(ɑ+β)
∵ ∠BFC = ∠D+∠DBE+∠EBF = ɑ+ɑ+2β = 2(ɑ+β)(外角定理和④的结论)
∴ ∠BFC = ∠BHC = ∠FCH = 2(ɑ+β)
∴∠ACH = 180˚-∠FCH = 180˚-∠BHC =∠AHC
∴∆AHC为等腰三角形,AH = AC = b = 2
∴ BH = BF =AB-AH = c-b = 1
⑦ 过B点做BG⊥AD,垂足为G。
∵∆ABD为等腰∆,∴ BG⊥且平分AD,DG = AG = AD/2
设FC的长度为x,AD = DF+FC+AC = c+b+x = 5+x, AG = (5+x)/2
⑧ 在Rt∆ABG中,AB = c = 3, AG = (5+x)/2
∴ BG² = AB²-AG² = 9-(5+x)²/4
⑨ 在Rt∆BGF中,BF = 1(见⑥的结论)
GF = AG-FC-AC = [(5+x)/2]-x-2 = (1-x)/2
BG² =BF²-GF² = 1-[(1-x)²/4]
⑩ 在Rt∆BGC中, BC = a, GC = AG-AC=[(5+x)/2]-2 = (1+x)/2
BG² =BC²-GC² = a² -[(1+x)²/4]
⑪ 对比⑧、⑨可得:9-(5+x)²/4 = 1-[(1-x)²/4],解方程可得:x = 2/3
⑫ 对比⑨、⑩和可得:a² -[(1+x)²/4] = 1-[(1-x)²/4],
化简后,a² = 1+x = 5/3,a = √15/3
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